Worked examples · Sheet 3

Derivatives of composite functions

Theory used: Differentiation rules Derivatives of elementary functions

Exercise 1

Compute the derivative of $f(x) = (3x^2 + 1)^5$ and of $g(x) = \ln(x^2 + 4)$ .

f'(x) = 5\,(3x^2 + 1)^4 \cdot 6x = 30x\,(3x^2 + 1)^4

g'(x) = \frac{1}{x^2 + 4} \cdot 2x = \frac{2x}{x^2 + 4}

Both are composite functions: in $f$ the inner function is $3x^2 + 1$ and the outer one is the fifth power; in $g$ the inner function is $x^2 + 4$ and the outer one is the logarithm. By the composite function rule one differentiates the outer function leaving the inner one unchanged, and multiplies by the derivative of the inner function. For $f$ the power gives $5\,(3x^2 + 1)^4$ and the inner function has derivative $6x$ . For $g$ the logarithm gives $\dfrac{1}{x^2 + 4}$ and the inner function has derivative $2x$ . Expanding $(3x^2 + 1)^5$ before differentiating would give the same result with much more computation.

Exercise 2

Compute the derivative of $f(x) = \sqrt{1 - x^2}$ , together with its domain of differentiability.

f(x) = (1 - x^2)^{\frac{1}{2}}, \qquad f'(x) = \frac{1}{2}\,(1 - x^2)^{-\frac{1}{2}} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}}

-1 < x < 1

The root is written as a power with exponent $\dfrac{1}{2}$ , so the outer function is a power and the inner function is $1 - x^2$ . By the composite function rule the derivative of the power, $\dfrac{1}{2}\,(1 - x^2)^{-\frac{1}{2}}$ , is multiplied by $-2x$ , and canceling the $2$ gives $-\dfrac{x}{\sqrt{1 - x^2}}$ . The function is defined for $-1 \leq x \leq 1$ , but the derivative has the root in the denominator, so it requires $1 - x^2 > 0$ ; hence at the endpoints $\pm 1$ the function is not differentiable, and the graph, a semicircle, has a vertical tangent there.

Exercise 3

Compute the derivative of $f(x) = 2^{3x}$ and of $g(x) = x^2 e^{-x}$ .

f'(x) = 2^{3x} \ln 2 \cdot 3 = 3 \ln 2 \cdot 2^{3x}

g'(x) = 2x\,e^{-x} + x^2 \cdot \left( -e^{-x} \right) = x\,(2 - x)\,e^{-x}

In $f$ the outer function is the exponential with base $2$ , which in its variable $u$ has derivative $2^{u} \ln 2$ , and the inner function is $3x$ , with derivative $3$ . In $g$ the product rule and the composite function rule appear together; in particular the factor $e^{-x}$ is differentiated as a composite function and gives $e^{-x} \cdot (-1)$ . Factoring out $x\,e^{-x}$ gives the form $x\,(2 - x)\,e^{-x}$ , from which one reads that the derivative vanishes at $0$ and at $2$ .