Derivatives of inverse functions: ln x, arcsin x, arctan x

Warm-up

Three quick questions from earlier lessons.

Discover

$y = \ln x$ means $e^y = x$ . The slope of $e^y$ is $e^y = x$ . The graphs of $\ln$ and $\exp$ are reflections, so their slopes are reciprocal. What is $(\ln x)'$ ?

$\dfrac{1}{x}$
$\ln x$

Key idea

The derivatives of the inverse functions come from the chain rule.

(\ln x)' = \dfrac{1}{x}

(

x > 0

);

(\arcsin x)' = \dfrac{1}{\sqrt{1 - x^2}}

;

(\arctan x)' = \dfrac{1}{1 + x^2}

. With an inside function:

(\ln u)' = \dfrac{u'}{u}

.

Worked example

$h(x) = \ln(x^3 + 2)$ . Find $h'(x)$ .

(\ln u)' = \dfrac{u'}{u}

Inside u = x³ + 2, u' = 3x².

h'(x) = \dfrac{3x^2}{x^3 + 2}

Substitute.

Your turn, guided

Your turn: $h(x) = \arctan(x^2)$ .

h'(x) = \dfrac{2x}{1 + x^4}

u = x², u' = 2x; divide by 1 + u² = 1 + x⁴.

Practice ★

$f(x) = \ln(9x)$ for $x > 0$ . Find $f'(x)$ .
$f(x) = \ln(2x)$ for $x > 0$ . Find $f'(x)$ .

$f(x) = \ln(x^2 + 2)$ . Find $f'(x)$ .
$f(x) = \ln(x^2 + 8)$ . Find $f'(x)$ .

Watch out!

Practice ★★

$f(x) = \arctan(7x)$ . Find $f'(x)$ .
$f(x) = \arctan(2x)$ . Find $f'(x)$ .

Exit ticket

Three questions to check you have got it.

$f(x) = \ln(x^2 + 2)$ . Find $f'(x)$ .
$f(x) = \ln(x^2 + 8)$ . Find $f'(x)$ .
$f(x) = \ln(x^2 + 1)$ . Find $f'(x)$ .

Summary card

Inverse functions

$(\ln x)' = \dfrac{1}{x}$ , $(\ln u)' = \dfrac{u'}{u}$ .
$(\arcsin x)' = \dfrac{1}{\sqrt{1 - x^2}}$ , $(\arctan x)' = \dfrac{1}{1 + x^2}$ .

Watch out: the chain rule still applies to the inside.

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