Worked examples · Sheet 2

Derivatives of composite functions

Theory used: Derivative of a composite function Derivatives of the elementary functions

Exercise 1

Compute the derivative of $f (x) = sin (3 x^{2} + 1)$ .

f^{'} (x) = cos (3 x^{2} + 1) \cdot (3 x^{2} + 1)^{'} = 6 x cos (3 x^{2} + 1)

The function is composite: the outer function is the sine, the argument is $3 x^{2} + 1$ . Since the derivative of the sine is the cosine, one writes $cos (3 x^{2} + 1)$ leaving the argument unchanged; the chain rule then requires multiplying by the derivative of the argument, which equals $6 x$ . The derivative is therefore $6 x cos (3 x^{2} + 1)$ , defined on all of $R$ .

Exercise 2

Compute the derivative of $f (x) = ln (x^{2} + 4)$ .

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

The polynomial $x^{2} + 4$ is positive for every $x$ , therefore the function is defined on all of $R$ . The outer function is the logarithm, whose derivative is the reciprocal of the argument, hence the first factor is $\frac{1}{x ^{2} + 4}$ ; the derivative of the argument equals $2 x$ . The product of the two factors gives $f^{'} (x) = \frac{2 x}{x ^{2} + 4}$ .

Exercise 3

Compute the derivative of $f (x) = 1 + e^{2 x}$ .

f^{'} (x) = \frac{1}{2 1 + e ^{2 x}} \cdot (1 + e^{2 x})^{'} = \frac{2 e ^{2 x}}{2 1 + e ^{2 x}} = \frac{e ^{2 x}}{1 + e ^{2 x}}

The composition has three levels: the square root, the function $1 + e^{2 x}$ and the exponent $2 x$ . The derivative of the root is $\frac{1}{2 \cdot}$ with the argument unchanged, therefore the first factor is $\frac{1}{2 1 + e ^{2 x}}$ . To differentiate $1 + e^{2 x}$ one applies the chain rule again: the derivative of $e^{2 x}$ is $e^{2 x}$ multiplied by the derivative of the exponent, that is, $2 e^{2 x}$ . The factor $2$ cancels with the $2$ of the root, hence $f^{'} (x) = \frac{e ^{2 x}}{1 + e ^{2 x}}$ . Since $1 + e^{2 x} > 0$ for every $x$ , the derivative is defined on all of $R$ .