Inverse-tangent Derivative | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

Inverse tangent is an angle

Definition

What is meant by inverse tangent?

Model answer: $y=\tan^{-1}x$ means $\tan y=x$ , with principal range $-\pi/2<y<\pi/2$ .

Common mistake

\tan^{-1}x

and

\cot x

are different functions.

\cot x=1/\tan x

Derive the rule

Undo the inverse before differentiating

1. Inverse formy = tan⁻¹u

2. Ordinary formtan y = u

3. Differentiatesec²y · y′ = u′

4. Replacey′ = u′/(1+u²)

tan⁻¹ returns −π/2 < y < π/2

The denominator is one plus u squared; the chain-rule numerator is u prime.

y=\tan^{-1}u\quad\Longleftrightarrow\quad\tan y=u

Differentiate implicitly because $y$ is defined by the equation $\tan y=u(x)$ , not given directly as a function of $x$ .

In $\tan y=u(x)$ , the angle $y$ changes when $x$ changes. The chain rule therefore attaches $dy/dx$ when $\tan y$ is differentiated.

\sec^2y\,\frac{dy}{dx}=u'

Since $\sec^2y=1+\tan^2y=1+u^2$ ,

\frac{d}{dx}(\tan^{-1}u)=\frac{u'}{1+u^2}

The denominator is positive for real $u$ . The sign of the derivative therefore comes from $u'$ .

Combine rules

Worked example

A composite inverse tangent

y=\tan^{-1}(3x-1)

y'=\frac3{1+(3x-1)^2}

Your turnIndependent check [5 marks]

Differentiate $y=x\tan^{-1}(2x)$ , then evaluate $dy/dx$ at $x=1$ .

Show worked answer

y'=\tan^{-1}(2x)+\frac{2x}{1+4x^2}

Answer

\left.\frac{dy}{dx}\right|_{x=1}=\tan^{-1}2+\frac25