Basic Derivatives — Dynamic Practice
Basic Derivatives — Dynamic Practice. Practice questions to deepen understanding of basic derivatives. Online math practice with full solutions and clear explanations.
Dynamic practice in basic derivatives — differentiation rules, powers, sums, constants, and simple derivative calculations. New questions every attempt.
Question 1
7.69 pts
Find the derivative of the function:
\(f(x) = (x+2)^{2}\)
\(f(x) = (x+2)^{2}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = (x+2)^{2}\)
🧅 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.
Answer: \(f'(x) = 2(x+2)^{1} \cdot (1)\)
The function: \(f(x) = (x+2)^{2}\)
🧅 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.
| 🔵 Outer layer: | The power ^2 |
| 🟢 Inner layer: | \(x+2\) |
| 🟠 Inner derivative: | \(1\) |
Answer: \(f'(x) = 2(x+2)^{1} \cdot (1)\)
Question 2
7.69 pts
Find the derivative of the function:
\(f(x) = (3x-3)^{2}\)
\(f(x) = (3x-3)^{2}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = (3x-3)^{2}\)
🧅 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.
Answer: \(f'(x) = 2(3x-3)^{1} \cdot (3)\)
The function: \(f(x) = (3x-3)^{2}\)
🧅 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.
| 🔵 Outer layer: | The power ^2 |
| 🟢 Inner layer: | \(3x-3\) |
| 🟠 Inner derivative: | \(3\) |
Answer: \(f'(x) = 2(3x-3)^{1} \cdot (3)\)
Question 3
7.69 pts
Find the derivative of the function:
\(f(x) = \sqrt{2x+2}\)
\(f(x) = \sqrt{2x+2}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{2x+2}\)
🧅 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.
Answer: \(f'(x) = \frac{2}{2\sqrt{2x+2}}\)
The function: \(f(x) = \sqrt{2x+2}\)
🧅 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.
| 🔵 Outer layer: | The root √ |
| 🟢 Inner layer: | \(2x+2\) |
| 🟠 Inner derivative: | \(2\) |
Answer: \(f'(x) = \frac{2}{2\sqrt{2x+2}}\)
Question 4
7.69 pts
Find the derivative of the function:
\(f(x) = (4x^{2}-5x+1)^{3}\)
\(f(x) = (4x^{2}-5x+1)^{3}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = (4x^{2}-5x+1)^{3}\)
🧅 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.
Answer: \(f'(x) = 3(4x^{2}-5x+1)^{2} \cdot (8x-5)\)
The function: \(f(x) = (4x^{2}-5x+1)^{3}\)
🧅 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.
| 🔵 Outer layer: | The power ^3 |
| 🟢 Inner layer: | \(4x^{2}-5x+1\) |
| 🟠 Inner derivative: | \(8x-5\) |
Answer: \(f'(x) = 3(4x^{2}-5x+1)^{2} \cdot (8x-5)\)
Question 5
7.69 pts
Find the derivative of the function:
\(f(x) = \sqrt{4x^{2}+3x-5}\)
\(f(x) = \sqrt{4x^{2}+3x-5}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{4x^{2}+3x-5}\)
🧅 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.
Answer: \(f'(x) = \frac{8x+3}{2\sqrt{4x^{2}+3x-5}}\)
The function: \(f(x) = \sqrt{4x^{2}+3x-5}\)
🧅 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.
| 🔵 Outer layer: | The root √ |
| 🟢 Inner layer: | \(4x^{2}+3x-5\) |
| 🟠 Inner derivative: | \(8x+3\) |
Answer: \(f'(x) = \frac{8x+3}{2\sqrt{4x^{2}+3x-5}}\)
Question 6
7.69 pts
Find the derivative of the function:
\(f(x) = \sqrt{5x^{2}+x+3}\)
\(f(x) = \sqrt{5x^{2}+x+3}\)
Explanation:
Solution – Chain rule (onion method 🧅):
The function: \(f(x) = \sqrt{5x^{2}+x+3}\)
🧅 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.
Answer: \(f'(x) = \frac{10x+1}{2\sqrt{5x^{2}+x+3}}\)
The function: \(f(x) = \sqrt{5x^{2}+x+3}\)
🧅 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.
| 🔵 Outer layer: | The root √ |
| 🟢 Inner layer: | \(5x^{2}+x+3\) |
| 🟠 Inner derivative: | \(10x+1\) |
Answer: \(f'(x) = \frac{10x+1}{2\sqrt{5x^{2}+x+3}}\)
Question 7
7.69 pts
Find the derivative of the function:
\(f(x) = \sqrt{5x+5} \cdot (4x+4)\)
\(f(x) = \sqrt{5x+5} \cdot (4x+4)\)
Explanation:
Solution – Product rule:
The function: \(f(x) = \sqrt{5x+5} \cdot (4x+4)\)
✖️ Product rule:
Derivative of the first times the second, plus the first times the derivative of the second.
Answer: \(f'(x) = (\frac{5}{2\sqrt{5x+5}})(4x+4) + (\sqrt{5x+5})(4)\)
The function: \(f(x) = \sqrt{5x+5} \cdot (4x+4)\)
✖️ Product rule:
\((f \cdot g)' = f' \cdot g + f \cdot g'\)
| f = \(\sqrt{5x+5}\) | f' = \(\frac{5}{2\sqrt{5x+5}}\) |
| ✖ Multiply diagonally and add | |
| g = \(4x+4\) | g' = \(4\) |
Answer: \(f'(x) = (\frac{5}{2\sqrt{5x+5}})(4x+4) + (\sqrt{5x+5})(4)\)
Question 8
7.69 pts
Find the derivative of the function:
\(f(x) = \frac{8x^{2}-2x-4}{3x+3}\)
\(f(x) = \frac{8x^{2}-2x-4}{3x+3}\)
Explanation:
Solution – Quotient rule:
The function: \(f(x) = \frac{8x^{2}-2x-4}{3x+3}\)
📐 Quotient rule:
Derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, all divided by the denominator squared.
Answer: \(f'(x) = \frac{(16x-2)(3x+3) - (8x^{2}-2x-4)(3)}{(3x+3)^2}\)
The function: \(f(x) = \frac{8x^{2}-2x-4}{3x+3}\)
📐 Quotient rule:
\(\left(\frac{f}{g}\right)' = \frac{f' \cdot g - f \cdot g'}{g^2}\)
| f = \(8x^{2}-2x-4\) | f' = \(16x-2\) |
| ✖ Multiply diagonally | |
| g = \(3x+3\) | g' = \(3\) |
Answer: \(f'(x) = \frac{(16x-2)(3x+3) - (8x^{2}-2x-4)(3)}{(3x+3)^2}\)
Question 9
7.69 pts
Find the derivative of the function:
\(f(x) = \frac{3x^{2}-2x-3}{7x-6}\)
\(f(x) = \frac{3x^{2}-2x-3}{7x-6}\)
Explanation:
Solution – Quotient rule:
The function: \(f(x) = \frac{3x^{2}-2x-3}{7x-6}\)
📐 Quotient rule:
Derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, all divided by the denominator squared.
Answer: \(f'(x) = \frac{(6x-2)(7x-6) - (3x^{2}-2x-3)(7)}{(7x-6)^2}\)
The function: \(f(x) = \frac{3x^{2}-2x-3}{7x-6}\)
📐 Quotient rule:
\(\left(\frac{f}{g}\right)' = \frac{f' \cdot g - f \cdot g'}{g^2}\)
| f = \(3x^{2}-2x-3\) | f' = \(6x-2\) |
| ✖ Multiply diagonally | |
| g = \(7x-6\) | g' = \(7\) |
Answer: \(f'(x) = \frac{(6x-2)(7x-6) - (3x^{2}-2x-3)(7)}{(7x-6)^2}\)
Question 10
7.69 pts
Find the derivative of the function:
\(f(x) = (8x^{2}+5x+2) \cdot (6x+3)\)
\(f(x) = (8x^{2}+5x+2) \cdot (6x+3)\)
Explanation:
Solution – Product rule:
The function: \(f(x) = (8x^{2}+5x+2) \cdot (6x+3)\)
✖️ Product rule:
Derivative of the first times the second, plus the first times the derivative of the second.
Answer: \(f'(x) = (16x+5)(6x+3) + (8x^{2}+5x+2)(6)\)
The function: \(f(x) = (8x^{2}+5x+2) \cdot (6x+3)\)
✖️ Product rule:
\((f \cdot g)' = f' \cdot g + f \cdot g'\)
| f = \(8x^{2}+5x+2\) | f' = \(16x+5\) |
| ✖ Multiply diagonally and add | |
| g = \(6x+3\) | g' = \(6\) |
Answer: \(f'(x) = (16x+5)(6x+3) + (8x^{2}+5x+2)(6)\)
Question 11
7.69 pts
Find the derivative of the function:
\(f(x) = \sqrt{3x+5} \cdot (7x+1)\)
\(f(x) = \sqrt{3x+5} \cdot (7x+1)\)
Explanation:
Solution – Product rule:
The function: \(f(x) = \sqrt{3x+5} \cdot (7x+1)\)
✖️ Product rule:
Derivative of the first times the second, plus the first times the derivative of the second.
Answer: \(f'(x) = (\frac{3}{2\sqrt{3x+5}})(7x+1) + (\sqrt{3x+5})(7)\)
The function: \(f(x) = \sqrt{3x+5} \cdot (7x+1)\)
✖️ Product rule:
\((f \cdot g)' = f' \cdot g + f \cdot g'\)
| f = \(\sqrt{3x+5}\) | f' = \(\frac{3}{2\sqrt{3x+5}}\) |
| ✖ Multiply diagonally and add | |
| g = \(7x+1\) | g' = \(7\) |
Answer: \(f'(x) = (\frac{3}{2\sqrt{3x+5}})(7x+1) + (\sqrt{3x+5})(7)\)
Question 12
7.69 pts
Find the derivative of the function:
\(f(x) = \frac{\sqrt{2x-5}}{3x-5}\)
\(f(x) = \frac{\sqrt{2x-5}}{3x-5}\)
Explanation:
Solution – Quotient rule:
The function: \(f(x) = \frac{\sqrt{2x-5}}{3x-5}\)
📐 Quotient rule:
Derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, all divided by the denominator squared.
Answer: \(f'(x) = \frac{(\frac{2}{2\sqrt{2x-5}})(3x-5) - (\sqrt{2x-5})(3)}{(3x-5)^2}\)
The function: \(f(x) = \frac{\sqrt{2x-5}}{3x-5}\)
📐 Quotient rule:
\(\left(\frac{f}{g}\right)' = \frac{f' \cdot g - f \cdot g'}{g^2}\)
| f = \(\sqrt{2x-5}\) | f' = \(\frac{2}{2\sqrt{2x-5}}\) |
| ✖ Multiply diagonally | |
| g = \(3x-5\) | g' = \(3\) |
Answer: \(f'(x) = \frac{(\frac{2}{2\sqrt{2x-5}})(3x-5) - (\sqrt{2x-5})(3)}{(3x-5)^2}\)
Question 13
7.69 pts
Find the derivative of the function:
\(f(x) = \frac{\sqrt{6x+4}}{6x+1}\)
\(f(x) = \frac{\sqrt{6x+4}}{6x+1}\)
Explanation:
Solution – Quotient rule:
The function: \(f(x) = \frac{\sqrt{6x+4}}{6x+1}\)
📐 Quotient rule:
Derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, all divided by the denominator squared.
Answer: \(f'(x) = \frac{(\frac{6}{2\sqrt{6x+4}})(6x+1) - (\sqrt{6x+4})(6)}{(6x+1)^2}\)
The function: \(f(x) = \frac{\sqrt{6x+4}}{6x+1}\)
📐 Quotient rule:
\(\left(\frac{f}{g}\right)' = \frac{f' \cdot g - f \cdot g'}{g^2}\)
| f = \(\sqrt{6x+4}\) | f' = \(\frac{6}{2\sqrt{6x+4}}\) |
| ✖ Multiply diagonally | |
| g = \(6x+1\) | g' = \(6\) |
Answer: \(f'(x) = \frac{(\frac{6}{2\sqrt{6x+4}})(6x+1) - (\sqrt{6x+4})(6)}{(6x+1)^2}\)