Substitution method

Some inverse-trigonometric expressions can in principle be differentiated directly by repeated chain-rule and quotient-rule work, but that route is often extremely laborious. The substitution method replaces the algebra with a well-chosen trigonometric substitution that collapses the expression to something much simpler before any differentiation is attempted.

The idea, illustrated. Consider $f(x)=\tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)$. Differentiating this directly via the chain and quotient rules is possible but heavy. Instead, substitute $x=\tan\theta$. Then, recognising the double-angle identity for tangent,

so $f(x) = \tan^{-1}(\tan2\theta) = 2\theta = 2\tan^{-1}x$ (using $\theta=\tan^{-1}x$ to undo the substitution). Now differentiating this much simpler form is immediate: $f'(x) = 2\cdot\dfrac{1}{1+x^2} = \dfrac{2}{1+x^2}$.

Worked illustration — $\tan^{-1}\!\left(\dfrac{1+x}{1-x}\right)$. Substitute $x=\tan\theta$. Then $\dfrac{1+x}{1-x} = \dfrac{1+\tan\theta}{1-\tan\theta} = \tan\!\left(\dfrac\pi4+\theta\right)$ (the tangent-addition identity, since $\tan\tfrac\pi4=1$), so

and since $\pi/4$ is a constant, $y' = \dfrac{1}{1+x^2}$ directly — the substitution converts an intractable-looking quotient inside an inverse tangent into "a constant plus $\tan^{-1}x$", whose derivative is read off instantly.

Worked illustration — a cosine-of-inverse-cosine case. For $f(x)=\cos^{-1}\!\left(\dfrac{4x^3-3x}{1}\right)$-type expressions, substitute $x=\cos\theta$ and use the triple-angle identity $4\cos^3\theta-3\cos\theta=\cos3\theta$: e.g. for $f(x)=\cos^{-1}(4x^3-3x)$, with $x=\cos\theta$, $4x^3-3x=\cos3\theta$, so $f(x)=\cos^{-1}(\cos3\theta)=3\theta=3\cos^{-1}x$, giving $f'(x) = -\dfrac{3}{\sqrt{1-x^2}}$ immediately, in place of a direct chain-rule computation through a cubic inside an inverse cosine. …