3.4, 3.5 - Inverse tangent in calculus | Cambridge A-Level Mathematics

Inverse tangent connects a familiar angle function to a new kind of rational integral. In this lesson, 3.4 supplies its derivative and the chain rule extends it to composite functions; 3.5 reverses that derivative to integrate positive quadratic denominators. Principal radian values will then give exact definite integrals.

Differentiating inverse tangent

Which function has derivative $1/(1+x^2)$ ? It is not a power of $1+x^2$ : differentiating such a power introduces a factor of $x$ . The answer comes from reversing the tangent function.

Inverse tangent

$y=\tan^{-1}x$ , also written $y=\arctan x$ , means that $\tan y=x$ and

-\frac\pi2<y<\frac\pi2.

The inverse tangent takes any real input and returns the unique angle in this principal interval. Here, and throughout calculus with trigonometric functions, angles are in radians.

For example, $\tan^{-1}1=\pi/4$ and $\tan^{-1}(-1)=-\pi/4$ . Although $5\pi/4$ also has tangent $1$ , it is not the principal inverse-tangent value. The notation $\tan^{-1}x$ means an inverse function, whereas $(\tan x)^{-1}=\cot x$ means a reciprocal.

To find the derivative, start with $\tan y=x$ . Differentiating with respect to $x$ and retaining the chain-rule factor gives

\sec^2 y\,\frac{\mathrm dy}{\mathrm dx}=1.

Since $\sec^2y=1+\tan^2y=1+x^2$ , divide by this positive quantity.

Inverse-tangent derivative

\frac{\mathrm d}{\mathrm dx}\tan^{-1}x=\frac1{1+x^2},\qquad x\in\mathbb R.

In the formula booklet. For a differentiable inside function $u=u(x)$ , the chain rule therefore gives

\frac{\mathrm d}{\mathrm dx}\tan^{-1}u=\frac{u'}{1+u^2}.

The composite form follows from the supplied derivative.

For $y=\tan^{-1}x$ , the derivative is always positive but tends towards zero for large positive or negative inputs. The graph rises throughout, flattening towards its horizontal asymptotes.

[DIAGRAM: asset_name: An angle function and its gradient; asset_slug: inverse-tangent-gradient; caption: Inverse tangent rises between two horizontal asymptotes, with a positive gradient that decreases away from zero; description: Two coordinate panels share horizontal inputs from minus four to four. The left shows y equals inverse tangent x, points minus one comma minus pi over four, zero comma zero, and one comma pi over four, with dashed horizontal asymptotes y equals plus and minus pi over two. The right shows its gradient 1 over 1 plus x squared, with maximum one at zero and value one half at both minus one and one.]
Inverse tangent rises between two horizontal asymptotes, with a positive gradient that decreases away from zero

Worked example 1: a linear input

Question. Differentiate $y=\tan^{-1}(2x)$ .

Plan. Use the inverse-tangent derivative with inside function $u=2x$ , then multiply by $u'=2$ .

\frac{\mathrm dy}{\mathrm dx}=\frac{2}{1+(2x)^2}=\frac2{1+4x^2}.

Answer. $2/(1+4x^2)$ , for all real $x$ .

Check. At $x=0$ , the central difference $[\tan^{-1}(0.002)-\tan^{-1}(-0.002)]/0.002$ is approximately $2.00$ , agreeing with the derivative value $2$ .

Change only the input from $2x$ to $2x^2$ . Its derivative changes, and the whole input still has to be squared in the denominator.

Worked example 2: a quadratic input

Question. Differentiate $y=\tan^{-1}(2x^2)$ .

Plan. Use $u=2x^2$ , so $u'=4x$ , in the same chain-rule form.

\frac{\mathrm dy}{\mathrm dx}=\frac{4x}{1+(2x^2)^2}=\frac{4x}{1+4x^4}.

Answer. $4x/(1+4x^4)$ , for all real $x$ .

Check. The original function is even, so its gradient changes sign under $x\mapsto-x$ , as this odd derivative does. At $x=1$ , a central difference with step $0.001$ is approximately $0.800001$ , agreeing with $4/5$ .

Constant multiples and sums use the usual rules. For example, the derivative of $3\tan^{-1}x-\tan^{-1}(2x)$ is $3/(1+x^2)-2/(1+4x^2)$ . Products and quotients can combine this new derivative with the rules already learnt.

Watch out

For $\tan^{-1}(2x^2)$ , the denominator is $1+(2x^2)^2=1+4x^4$ , not $1+2x^2$ . Keep the entire inside expression in brackets before squaring it.

Differentiate another quadratic input before evaluating its gradient.

An inverse tangent integral

Reversing the basic derivative gives $\int1/(1+x^2)\,\mathrm dx=\tan^{-1}x+C$ . A different positive constant in the denominator requires a scale factor inside the inverse tangent and a matching factor outside.

For $a>0$ , differentiate $\tan^{-1}(x/a)$ :

\frac{\mathrm d}{\mathrm dx}\tan^{-1}\left(\frac xa\right) =\frac{1/a}{1+x^2/a^2}=\frac a{x^2+a^2}.

Dividing that result by $a$ gives the desired integrand.

A positive quadratic denominator

For a positive constant $a$ ,

\int\frac1{x^2+a^2}\,\mathrm dx =\frac1a\tan^{-1}\left(\frac xa\right)+C.

In the formula booklet, with the arbitrary constant omitted there. The denominator is positive for every real $x$ .

This is the sum-of-squares pattern. A difference such as $x^2-a^2$ does not have this antiderivative. Also distinguish $a^2$ in the denominator from $a$ in the outside factor and inside quotient.

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3.4, 3.5 - Inverse tangent in calculus

Differentiating inverse tangent

An inverse tangent integral

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