Derivative of √(trigonometric) — Dynamic Practice
Derivative of √(trigonometric) — Dynamic Practice. Practice questions to deepen understanding of derivatives where a trigonometric expression appears under a square root. Online math practice with full solutions and clear explanations.
Dynamic practice in derivatives of √(trigonometric expression) — combines the chain rule for the root with derivatives of trig functions.
Question 1
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(x+7)}\)
\(f(x) = \sqrt{\cos(x+7)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(x+7)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-\sin(x+7)}{2\sqrt{\cos(x+7)}}\)
The function: \(f(x) = \sqrt{\cos(x+7)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-\sin(x+7)}{2\sqrt{\cos(x+7)}}\)
Question 2
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(3x-5)}\)
\(f(x) = \sqrt{\sin(3x-5)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(3x-5)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{3\cos(3x-5)}{2\sqrt{\sin(3x-5)}}\)
The function: \(f(x) = \sqrt{\sin(3x-5)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{3\cos(3x-5)}{2\sqrt{\sin(3x-5)}}\)
Question 3
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(6x+9)}\)
\(f(x) = \sqrt{\cos(6x+9)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(6x+9)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-6\sin(6x+9)}{2\sqrt{\cos(6x+9)}}\)
The function: \(f(x) = \sqrt{\cos(6x+9)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-6\sin(6x+9)}{2\sqrt{\cos(6x+9)}}\)
Question 4
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(x+2)}\)
\(f(x) = \sqrt{\sin(x+2)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(x+2)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{\cos(x+2)}{2\sqrt{\sin(x+2)}}\)
The function: \(f(x) = \sqrt{\sin(x+2)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{\cos(x+2)}{2\sqrt{\sin(x+2)}}\)
Question 5
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(4x-9)}\)
\(f(x) = \sqrt{\sin(4x-9)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(4x-9)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{4\cos(4x-9)}{2\sqrt{\sin(4x-9)}}\)
The function: \(f(x) = \sqrt{\sin(4x-9)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{4\cos(4x-9)}{2\sqrt{\sin(4x-9)}}\)
Question 6
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(4x-6)}\)
\(f(x) = \sqrt{\sin(4x-6)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(4x-6)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{4\cos(4x-6)}{2\sqrt{\sin(4x-6)}}\)
The function: \(f(x) = \sqrt{\sin(4x-6)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{4\cos(4x-6)}{2\sqrt{\sin(4x-6)}}\)
Question 7
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(6x-2)}\)
\(f(x) = \sqrt{\sin(6x-2)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(6x-2)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{6\cos(6x-2)}{2\sqrt{\sin(6x-2)}}\)
The function: \(f(x) = \sqrt{\sin(6x-2)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{6\cos(6x-2)}{2\sqrt{\sin(6x-2)}}\)
Question 8
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(5x-7)}\)
\(f(x) = \sqrt{\sin(5x-7)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(5x-7)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{5\cos(5x-7)}{2\sqrt{\sin(5x-7)}}\)
The function: \(f(x) = \sqrt{\sin(5x-7)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{5\cos(5x-7)}{2\sqrt{\sin(5x-7)}}\)
Question 9
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(x)}\)
\(f(x) = \sqrt{\sin(x)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(x)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{\cos(x)}{2\sqrt{\sin(x)}}\)
The function: \(f(x) = \sqrt{\sin(x)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{\cos(x)}{2\sqrt{\sin(x)}}\)
Question 10
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(7x+4)}\)
\(f(x) = \sqrt{\sin(7x+4)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(7x+4)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{7\cos(7x+4)}{2\sqrt{\sin(7x+4)}}\)
The function: \(f(x) = \sqrt{\sin(7x+4)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{7\cos(7x+4)}{2\sqrt{\sin(7x+4)}}\)
Question 11
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(3x+10)}\)
\(f(x) = \sqrt{\cos(3x+10)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(3x+10)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-3\sin(3x+10)}{2\sqrt{\cos(3x+10)}}\)
The function: \(f(x) = \sqrt{\cos(3x+10)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-3\sin(3x+10)}{2\sqrt{\cos(3x+10)}}\)
Question 12
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(4x)}\)
\(f(x) = \sqrt{\sin(4x)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(4x)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{4\cos(4x)}{2\sqrt{\sin(4x)}}\)
The function: \(f(x) = \sqrt{\sin(4x)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{4\cos(4x)}{2\sqrt{\sin(4x)}}\)
Question 13
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(5x-2)}\)
\(f(x) = \sqrt{\cos(5x-2)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(5x-2)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-5\sin(5x-2)}{2\sqrt{\cos(5x-2)}}\)
The function: \(f(x) = \sqrt{\cos(5x-2)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-5\sin(5x-2)}{2\sqrt{\cos(5x-2)}}\)
Question 14
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(3x-10)}\)
\(f(x) = \sqrt{\cos(3x-10)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(3x-10)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-3\sin(3x-10)}{2\sqrt{\cos(3x-10)}}\)
The function: \(f(x) = \sqrt{\cos(3x-10)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-3\sin(3x-10)}{2\sqrt{\cos(3x-10)}}\)
Question 15
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(6x+10)}\)
\(f(x) = \sqrt{\sin(6x+10)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(6x+10)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{6\cos(6x+10)}{2\sqrt{\sin(6x+10)}}\)
The function: \(f(x) = \sqrt{\sin(6x+10)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{6\cos(6x+10)}{2\sqrt{\sin(6x+10)}}\)
Question 16
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(2x+11)}\)
\(f(x) = \sqrt{\sin(2x+11)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(2x+11)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{2\cos(2x+11)}{2\sqrt{\sin(2x+11)}}\)
The function: \(f(x) = \sqrt{\sin(2x+11)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{2\cos(2x+11)}{2\sqrt{\sin(2x+11)}}\)
Question 17
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(x+11)}\)
\(f(x) = \sqrt{\sin(x+11)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(x+11)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{\cos(x+11)}{2\sqrt{\sin(x+11)}}\)
The function: \(f(x) = \sqrt{\sin(x+11)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{\cos(x+11)}{2\sqrt{\sin(x+11)}}\)
Question 18
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(4x+4)}\)
\(f(x) = \sqrt{\sin(4x+4)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(4x+4)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{4\cos(4x+4)}{2\sqrt{\sin(4x+4)}}\)
The function: \(f(x) = \sqrt{\sin(4x+4)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{4\cos(4x+4)}{2\sqrt{\sin(4x+4)}}\)
Question 19
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(8x+1)}\)
\(f(x) = \sqrt{\cos(8x+1)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(8x+1)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-8\sin(8x+1)}{2\sqrt{\cos(8x+1)}}\)
The function: \(f(x) = \sqrt{\cos(8x+1)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-8\sin(8x+1)}{2\sqrt{\cos(8x+1)}}\)
Question 20
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(x-7)}\)
\(f(x) = \sqrt{\cos(x-7)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(x-7)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-\sin(x-7)}{2\sqrt{\cos(x-7)}}\)
The function: \(f(x) = \sqrt{\cos(x-7)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-\sin(x-7)}{2\sqrt{\cos(x-7)}}\)
Question 21
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(5x+9)}\)
\(f(x) = \sqrt{\sin(5x+9)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(5x+9)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{5\cos(5x+9)}{2\sqrt{\sin(5x+9)}}\)
The function: \(f(x) = \sqrt{\sin(5x+9)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{5\cos(5x+9)}{2\sqrt{\sin(5x+9)}}\)
Question 22
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(x-9)}\)
\(f(x) = \sqrt{\cos(x-9)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(x-9)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-\sin(x-9)}{2\sqrt{\cos(x-9)}}\)
The function: \(f(x) = \sqrt{\cos(x-9)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-\sin(x-9)}{2\sqrt{\cos(x-9)}}\)
Question 23
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(x-1)}\)
\(f(x) = \sqrt{\sin(x-1)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(x-1)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{\cos(x-1)}{2\sqrt{\sin(x-1)}}\)
The function: \(f(x) = \sqrt{\sin(x-1)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{\cos(x-1)}{2\sqrt{\sin(x-1)}}\)
Question 24
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(8x-1)}\)
\(f(x) = \sqrt{\cos(8x-1)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(8x-1)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-8\sin(8x-1)}{2\sqrt{\cos(8x-1)}}\)
The function: \(f(x) = \sqrt{\cos(8x-1)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-8\sin(8x-1)}{2\sqrt{\cos(8x-1)}}\)
Question 25
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(2x-3)}\)
\(f(x) = \sqrt{\sin(2x-3)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(2x-3)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{2\cos(2x-3)}{2\sqrt{\sin(2x-3)}}\)
The function: \(f(x) = \sqrt{\sin(2x-3)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{2\cos(2x-3)}{2\sqrt{\sin(2x-3)}}\)
Question 26
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(7x+2)}\)
\(f(x) = \sqrt{\sin(7x+2)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(7x+2)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{7\cos(7x+2)}{2\sqrt{\sin(7x+2)}}\)
The function: \(f(x) = \sqrt{\sin(7x+2)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{7\cos(7x+2)}{2\sqrt{\sin(7x+2)}}\)
Question 27
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(8x)}\)
\(f(x) = \sqrt{\cos(8x)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(8x)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-8\sin(8x)}{2\sqrt{\cos(8x)}}\)
The function: \(f(x) = \sqrt{\cos(8x)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-8\sin(8x)}{2\sqrt{\cos(8x)}}\)
Question 28
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(6x+4)}\)
\(f(x) = \sqrt{\cos(6x+4)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(6x+4)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-6\sin(6x+4)}{2\sqrt{\cos(6x+4)}}\)
The function: \(f(x) = \sqrt{\cos(6x+4)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-6\sin(6x+4)}{2\sqrt{\cos(6x+4)}}\)
Question 29
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(x-1)}\)
\(f(x) = \sqrt{\cos(x-1)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(x-1)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-\sin(x-1)}{2\sqrt{\cos(x-1)}}\)
The function: \(f(x) = \sqrt{\cos(x-1)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-\sin(x-1)}{2\sqrt{\cos(x-1)}}\)
Question 30
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(x)}\)
\(f(x) = \sqrt{\cos(x)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(x)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-\sin(x)}{2\sqrt{\cos(x)}}\)
The function: \(f(x) = \sqrt{\cos(x)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-\sin(x)}{2\sqrt{\cos(x)}}\)
Question 31
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(7x-12)}\)
\(f(x) = \sqrt{\sin(7x-12)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(7x-12)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{7\cos(7x-12)}{2\sqrt{\sin(7x-12)}}\)
The function: \(f(x) = \sqrt{\sin(7x-12)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{7\cos(7x-12)}{2\sqrt{\sin(7x-12)}}\)
Question 32
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(8x+7)}\)
\(f(x) = \sqrt{\sin(8x+7)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(8x+7)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{8\cos(8x+7)}{2\sqrt{\sin(8x+7)}}\)
The function: \(f(x) = \sqrt{\sin(8x+7)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{8\cos(8x+7)}{2\sqrt{\sin(8x+7)}}\)
Question 33
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(8x+2)}\)
\(f(x) = \sqrt{\sin(8x+2)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(8x+2)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{8\cos(8x+2)}{2\sqrt{\sin(8x+2)}}\)
The function: \(f(x) = \sqrt{\sin(8x+2)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{8\cos(8x+2)}{2\sqrt{\sin(8x+2)}}\)
Question 34
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(3x-8)}\)
\(f(x) = \sqrt{\sin(3x-8)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(3x-8)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{3\cos(3x-8)}{2\sqrt{\sin(3x-8)}}\)
The function: \(f(x) = \sqrt{\sin(3x-8)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{3\cos(3x-8)}{2\sqrt{\sin(3x-8)}}\)
Question 35
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(x-11)}\)
\(f(x) = \sqrt{\cos(x-11)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(x-11)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-\sin(x-11)}{2\sqrt{\cos(x-11)}}\)
The function: \(f(x) = \sqrt{\cos(x-11)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-\sin(x-11)}{2\sqrt{\cos(x-11)}}\)
Question 36
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(8x+7)}\)
\(f(x) = \sqrt{\cos(8x+7)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(8x+7)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-8\sin(8x+7)}{2\sqrt{\cos(8x+7)}}\)
The function: \(f(x) = \sqrt{\cos(8x+7)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-8\sin(8x+7)}{2\sqrt{\cos(8x+7)}}\)
Question 37
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(4x-5)}\)
\(f(x) = \sqrt{\sin(4x-5)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(4x-5)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{4\cos(4x-5)}{2\sqrt{\sin(4x-5)}}\)
The function: \(f(x) = \sqrt{\sin(4x-5)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{4\cos(4x-5)}{2\sqrt{\sin(4x-5)}}\)
Question 38
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\sin(4x-10)}\)
\(f(x) = \sqrt{\sin(4x-10)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\sin(4x-10)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{4\cos(4x-10)}{2\sqrt{\sin(4x-10)}}\)
The function: \(f(x) = \sqrt{\sin(4x-10)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{4\cos(4x-10)}{2\sqrt{\sin(4x-10)}}\)
Question 39
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(x-5)}\)
\(f(x) = \sqrt{\cos(x-5)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(x-5)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-\sin(x-5)}{2\sqrt{\cos(x-5)}}\)
The function: \(f(x) = \sqrt{\cos(x-5)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-\sin(x-5)}{2\sqrt{\cos(x-5)}}\)
Question 40
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(x-8)}\)
\(f(x) = \sqrt{\cos(x-8)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(x-8)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-\sin(x-8)}{2\sqrt{\cos(x-8)}}\)
The function: \(f(x) = \sqrt{\cos(x-8)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-\sin(x-8)}{2\sqrt{\cos(x-8)}}\)
Question 41
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(5x)}\)
\(f(x) = \sqrt{\cos(5x)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(5x)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-5\sin(5x)}{2\sqrt{\cos(5x)}}\)
The function: \(f(x) = \sqrt{\cos(5x)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-5\sin(5x)}{2\sqrt{\cos(5x)}}\)
Question 42
2.38 pts
Find the derivative of the function:
\(f(x) = \sqrt{\cos(3x+3)}\)
\(f(x) = \sqrt{\cos(3x+3)}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{\cos(3x+3)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
Answer: \(f'(x) = \frac{-3\sin(3x+3)}{2\sqrt{\cos(3x+3)}}\)
The function: \(f(x) = \sqrt{\cos(3x+3)}\)
🧅 Onion method:
Look at the function as layers, like an onion.
Start from the outer layer, differentiate it,
then multiply by the derivative of what is inside.
📊 Trigonometric derivatives:
| \((\sin x)' = \cos x\) | \((\cos x)' = -\sin x\) |
| \((\tan x)' = \frac{1}{\cos^2 x}\) | \((\cot x)' = \frac{-1}{\sin^2 x}\) |
Answer: \(f'(x) = \frac{-3\sin(3x+3)}{2\sqrt{\cos(3x+3)}}\)