Derivative of Square Root Functions

Derivative of Square Root Functions

Square root functions appear in a wide range of calculus problems. To differentiate them we use the basic square root derivative formula together with the chain rule, which handles the expression under the radical. On this page we develop the intuition behind the formula and practice evaluating the derivative at a point.

Background and Basic Definitions

A square root is simply a power of one-half: \(\sqrt{x}=x^{1/2}\). The basic derivative is therefore:

\[ \big(\sqrt{x}\big)' = \frac{1}{2\sqrt{x}} \]

When the expression under the radical is a full expression rather than just \(x\), we apply the chain rule: the outer function is the square root and the inner function is the expression \(g(x)\). This gives the general formula:

\[ \big(\sqrt{g(x)}\big)' = \frac{g'(x)}{2\sqrt{g(x)}} \]

The intuition: differentiate the square root as usual (\(\tfrac{1}{2\sqrt{\;}}\)) and then multiply by the derivative of the inner expression \(g'(x)\), exactly as with any composite function.

Recall: \((ax+b)'=a\) and \((x^2+c)'=2x\). To evaluate the derivative at a point — first find the derivative expression, substitute \(x\), and finally compute the square root in the denominator.

Basic Derivatives Table:

FunctionDerivativeFunctionDerivative
\(x^t\)\(t\cdot x^{t-1}\)\(\frac{1}{x}\)\(-\frac{1}{x^2}\)
\(\sqrt{x}\)\(\frac{1}{2\sqrt{x}}\)\(a^x\)\(a^x\cdot \ln a\)
\(\sin x\)\(\cos x\)\(\cos x\)\(-\sin x\)
\(\tan x\)\(\frac{1}{\cos^2 x}\)\(\log_a x\)\(\frac{1}{x\cdot \ln a}\)

where \(t\) is any real number. Differentiation rules: product \([f\cdot g]'=f'g+fg'\); quotient \(\left[\frac{f}{g}\right]'=\frac{f'g-fg'}{[g]^2}\); chain rule \([f(u(x))]'=f'(u)\cdot u'(x)\).

Solution Steps

  1. Step 1 — Identify the expression under the radical; that is the inner function \(g(x)\).
  2. Step 2 — Differentiate the inner function to get \(g'(x)\).
  3. Step 3 — Write the derivative using the formula \(\dfrac{g'(x)}{2\sqrt{g(x)}}\).
  4. Step 4 — If \(f'(a)\) is requested, substitute \(a\) into \(g(x)\) under the radical and into \(g'(x)\) in the numerator.
  5. Step 5 — Compute the square root in the denominator and simplify the fraction.
  6. Step 6 — Check that the expression under the radical is positive at the point; if it is zero, the derivative is undefined there.

Worked Examples

Example 1: Basic Square Root — Derivative at a Point

Problem: Given the function \(f(x)=\sqrt{x}\). Find \(f'(9)\).

Solution:

  1. Basic derivative of the square root: \(f'(x)=\dfrac{1}{2\sqrt{x}}\).
  2. Substitute \(x=9\): \(f'(9)=\dfrac{1}{2\sqrt{9}}\).
  3. Compute the square root: \(\sqrt{9}=3\).
  4. Finish: \(f'(9)=\dfrac{1}{2\cdot 3}=\dfrac{1}{6}\).

Answer: \(f'(9)=\dfrac{1}{6}\).

Example 2: Square Root of a Linear Expression

Problem: Given the function \(f(x)=\sqrt{2x}\). Find \(f'(8)\).

Solution:

  1. The inner function is \(g(x)=2x\), so \(g'(x)=2\).
  2. By the chain rule: \(f'(x)=\dfrac{2}{2\sqrt{2x}}=\dfrac{1}{\sqrt{2x}}\).
  3. Substitute \(x=8\): \(f'(8)=\dfrac{1}{\sqrt{2\cdot 8}}=\dfrac{1}{\sqrt{16}}\).
  4. Compute: \(\sqrt{16}=4\), so \(f'(8)=\dfrac{1}{4}\).

Answer: \(f'(8)=\dfrac{1}{4}\).

Example 3: Square Root of a General Linear Expression

Problem: Given the function \(f(x)=\sqrt{4x+1}\). Find \(f'(2)\).

Solution:

  1. The inner function is \(g(x)=4x+1\), so \(g'(x)=4\).
  2. By the chain rule: \(f'(x)=\dfrac{4}{2\sqrt{4x+1}}=\dfrac{2}{\sqrt{4x+1}}\).
  3. Substitute \(x=2\): \(f'(2)=\dfrac{2}{\sqrt{4\cdot 2+1}}=\dfrac{2}{\sqrt{9}}\).
  4. Compute: \(\sqrt{9}=3\), so \(f'(2)=\dfrac{2}{3}\).

Answer: \(f'(2)=\dfrac{2}{3}\).

Example 4: Square Root of a Quadratic Expression

Problem: Given the function \(f(x)=\sqrt{x^2+9}\). Find \(f'(4)\).

Solution:

  1. The inner function is \(g(x)=x^2+9\), so \(g'(x)=2x\).
  2. By the chain rule: \(f'(x)=\dfrac{2x}{2\sqrt{x^2+9}}=\dfrac{x}{\sqrt{x^2+9}}\).
  3. Substitute \(x=4\): \(f'(4)=\dfrac{4}{\sqrt{4^2+9}}=\dfrac{4}{\sqrt{25}}\).
  4. Compute: \(\sqrt{25}=5\), so \(f'(4)=\dfrac{4}{5}\).

Answer: \(f'(4)=\dfrac{4}{5}\).

Example 5: Square Root with a Negative Inner Coefficient

Problem: Given the function \(f(x)=\sqrt{25-x}\). Find \(f'(9)\).

Solution:

  1. The inner function is \(g(x)=25-x\), so \(g'(x)=-1\).
  2. By the chain rule: \(f'(x)=\dfrac{-1}{2\sqrt{25-x}}\).
  3. Substitute \(x=9\): \(f'(9)=\dfrac{-1}{2\sqrt{25-9}}=\dfrac{-1}{2\sqrt{16}}\).
  4. Compute: \(\sqrt{16}=4\), so \(f'(9)=\dfrac{-1}{2\cdot 4}=-\dfrac{1}{8}\).
  5. The negative sign makes sense: the function is decreasing as \(x\) increases.

Answer: \(f'(9)=-\dfrac{1}{8}\).

Common Mistakes

✗ Common mistake: Differentiating the square root without the chain rule and writing \(\big(\sqrt{g(x)}\big)'=\dfrac{1}{2\sqrt{g(x)}}\) without the factor \(g'(x)\).

✓ The correct way: When the expression under the radical is not just \(x\), you must multiply by the derivative of the inner expression: \(\dfrac{g'(x)}{2\sqrt{g(x)}}\). For example \((\sqrt{4x+1})'=\dfrac{4}{2\sqrt{4x+1}}\).

✗ Common mistake: Forgetting the negative sign of \(g'(x)\) when the inner expression is decreasing, for example in \(\sqrt{25-x}\).

✓ The correct way: The derivative of the inner function \(25-x\) is \(-1\). The negative sign carries through to the numerator, making the derivative negative — consistent with the function decreasing.

✗ Common mistake: Computing the square root in the denominator before substituting, or substituting \(x\) only in the denominator and not in the numerator.

✓ The correct way: Substitute \(x\) into both the numerator (\(g'(x)\)) and the denominator (\(g(x)\)), then compute the square root and simplify.

Practice Tips

  • Tip — Think of a square root as a power of one-half: \(\sqrt{g}=g^{1/2}\). That explains where the \(\tfrac{1}{2}\) comes from and why the square root moves to the denominator.
  • Tip — Choose convenient points: if the expression under the radical is a perfect square (\(4,9,16,25\)), the square root is an integer and the computation is clean.
  • Tip — Always check that the expression under the radical is positive at the given point; if it is zero, the radical is in the denominator and the derivative is undefined.
  • Tip — The sign of the derivative tells you about the trend: positive means the function is increasing, negative means it is decreasing — a quick way to check your answer.

Summary and Key Formulas

Key formulas:

\[ \big(\sqrt{x}\big)' = \frac{1}{2\sqrt{x}}, \qquad \big(\sqrt{g(x)}\big)' = \frac{g'(x)}{2\sqrt{g(x)}} \]
  • Differentiate the square root and multiply by the derivative of the inner expression.
  • To find \(f'(a)\): substitute \(a\) into both numerator and denominator, then compute the square root.
  • The expression under the radical must be positive at the point.