Derivatives of Standard Inverse Trigonometric Composite Functions, Key Identities and Substitutions

The six inverse-trig derivatives generalise to a composite argument $f(x)$ by the chain rule (Table 1.2.2), exactly the way the ordinary trig functions generalised in section 1.1.3 — every formula picks up an extra factor of $f'(x)$. Table 1.2.3 lists the identities used throughout this section's examples to simplify an inverse-trig expression before differentiating (in particular the 'undoing' identities like $\sin^{-1}(\sin\theta)=\theta$, the complementary-angle identities, and the tangent addition/subtraction formulas), and Table 1.2.4 lists the standard trigonometric substitutions ($x=\sin\theta$, $x=\tan\theta$, $x=\sec\theta$, $x=a\cos2\theta$, etc.) that collapse a recurring surd or rational shape down to a single multiple angle. The general strategy in nearly every worked example below is: (1) recognise the shape, (2) substitute the matching trig variable, (3) use a double/triple-angle identity to reduce the whole expression to $\theta$ or a small multiple of it, (4) differentiate that simple result, (5) substitute the inverse substitution back to express the answer in $x$.

Worked Example 1. Prove $\sin^{-1}x+\cos^{-1}x=\dfrac\pi2$ using derivatives. Let $f(x)=\sin^{-1}x+\cos^{-1}x$. Differentiating, $f'(x)=\dfrac1{\sqrt{1-x^2}}-\dfrac1{\sqrt{1-x^2}}=0$, so $f(x)$ is a constant function. Evaluating at the convenient point $x=0$: $f(0)=\sin^{-1}0+\cos^{-1}0=0+\dfrac\pi2=\dfrac\pi2$. Since $f$ is constant and equals $\dfrac\pi2$ at one point, $f(x)=\dfrac\pi2$ for every $x$ in the domain — proving the identity.

Worked Example 2 — differentiate w.r.t. x:

1. $\sin^{-1}(x^3)$: $\dfrac{dy}{dx}=\dfrac{3x^2}{\sqrt{1-x^6}}$.
2. $\cos^{-1}(2x^2-x)$: writing $\cos y=2x^2-x$ and differentiating implicitly, $-\sin y\dfrac{dy}{dx}=4x-1$, so $\dfrac{dy}{dx}=\dfrac{1-4x}{\sin y}=\dfrac{1-4x}{\sqrt{1-(2x^2-x)^2}}$ (equivalently reached directly via the chain rule on $\cos^{-1}u$).
3. $\sin^{-1}(2x)$: $\dfrac{dy}{dx}=\dfrac{2^x\log2\cdot\ldots}{}$ — more precisely, differentiating $\sin^{-1}(2^x)$ (the base-$2$ exponential form used in this part) gives $\dfrac{dy}{dx}=\dfrac{2^x\log2}{\sqrt{1-4^x}}$.
4. $\cot^{-1}\dfrac1{x^2}=\tan^{-1}(x^2)$ (using the reciprocal identity): $\dfrac{dy}{dx}=\dfrac{2x}{1+x^4}$.
5. $\cos^{-1}\sqrt{\dfrac{1+x}{2}}$: this is the half-angle form of $\cos^{-1}x$, i.e. it equals $\dfrac12\cos^{-1}x$, so $\dfrac{dy}{dx}=-\dfrac1{2\sqrt{1-x^2}}$.
6. $\sin^2[\sin^{-1}(x^2)]$: since $\sin[\sin^{-1}u]=u$, this simplifies first to $(x^2)^2=x^4$, so $\dfrac{dy}{dx}=4x^3$. Worked Example 3 — differentiate w.r.t. x (each collapses via a triple- or double-angle identity to a linear function of $x$, so every derivative below is a constant):

(i) $\cos^{-1}(4\cos^3x-3\cos x)=\cos^{-1}(\cos3x)=3x$, so $\dfrac{dy}{dx}=3$.

(ii) $\cos^{-1}[\sin(4x)]=\cos^{-1}\!\left[\cos\!\left(\frac\pi2-4x\right)\right]=\frac\pi2-4x$, so $\dfrac{dy}{dx}=-4$.

(iii) $\sin^{-1}\sqrt{\dfrac{1-\cos x}{2}}=\sin^{-1}\!\left(\sin\dfrac x2\right)=\dfrac x2$, so $\dfrac{dy}{dx}=\dfrac12$.

(iv) $\tan^{-1}\dfrac{1-\cos3x}{\sin3x}=\tan^{-1}\!\left(\tan\dfrac{3x}2\right)=\dfrac{3x}2$, so $\dfrac{dy}{dx}=\dfrac32$.

(v) $\cot^{-1}\dfrac{\cos x}{1+\sin x}=\tan^{-1}\dfrac{1+\sin x}{\cos x}$, which (dividing numerator and denominator by $\cos\frac x2$ after a half-angle expansion) reduces to $\dfrac\pi4+\dfrac x2$, so $\dfrac{dy}{dx}=\dfrac12$.

Worked Example 4 — differentiate w.r.t. x (each is put into the form $\sin\alpha\cos x+\cos\alpha\sin x=\sin(x+\alpha)$ or similar, by matching the given coefficients to $\sin\alpha,\cos\alpha$ of a suitable auxiliary angle $\alpha=\tan^{-1}(\text{ratio})$, which is then a constant that vanishes on differentiating):

(i) $\sin^{-1}\dfrac{2\cos x+3\sin x}{\sqrt{13}}$: put $\frac2{\sqrt{13}}=\sin\alpha,\ \frac3{\sqrt{13}}=\cos\alpha$ (checked $\sin^2\alpha+\cos^2\alpha=1$); the expression becomes $\sin^{-1}[\sin(x+\alpha)]=x+\alpha$, a linear function of $x$ plus a constant, so $\dfrac{dy}{dx}=1$.

(ii) $\cos^{-1}\dfrac{3\sin x^2+4\cos x^2}{5}$: similarly reduces to $x^2-\tan^{-1}\frac34$, so $\dfrac{dy}{dx}=2x$.

(iii) $\sin^{-1}\dfrac{a\cos x-b\sin x}{\sqrt{a^2+b^2}}$: reduces to $\tan^{-1}\frac ab-x$, so $\dfrac{dy}{dx}=-1$.

Worked Example 5 — differentiate w.r.t. x (each uses a trig substitution from Table 1.2.4 and a double/triple-angle identity):

(i) $\sin^{-1}\dfrac{2x}{1+x^2}$: put $x=\tan\theta$; the expression becomes $\sin^{-1}(\sin2\theta)=2\theta=2\tan^{-1}x$, so $\dfrac{dy}{dx}=\dfrac2{1+x^2}$.

(ii) $\cos^{-1}(2x\sqrt{1-x^2})$: put $x=\sin\theta$; reduces to $\frac\pi2-2\sin^{-1}x$, so $\dfrac{dy}{dx}=-\dfrac2{\sqrt{1-x^2}}$.

(iii) $\text{cosec}^{-1}\dfrac1{3x-4x^3}=\sin^{-1}(3x-4x^3)$: put $x=\sin\theta$; reduces to $3\sin^{-1}x$, so $\dfrac{dy}{dx}=\dfrac3{\sqrt{1-x^2}}$.

(iv) $\tan^{-1}\dfrac{2e^x}{1-e^{2x}}$: put $e^x=\tan\theta$; reduces to $2\tan^{-1}(e^x)$, so $\dfrac{dy}{dx}=\dfrac{2e^x}{1+e^{2x}}$.

(v) $\cos^{-1}\dfrac{1-9x^2}{1+9x^2}$: put $3x=\tan\theta$; reduces to $2\tan^{-1}(3x)$, so $\dfrac{dy}{dx}=\dfrac6{1+9x^2}$.

(vi) $\cos^{-1}\dfrac{2^x-2^{-x}}{2^x+2^{-x}}$: put $2^x=\tan\theta$; reduces to $\pi-2\tan^{-1}(2^x)$, so $\dfrac{dy}{dx}=-\dfrac{2^{x+1}\log2}{1+2^{2x}}$.

(vii) $\tan^{-1}\sqrt{\dfrac{3-x}{3+x}}$: put $x=3\cos2\theta$; reduces to $\theta=\frac12\cos^{-1}\frac x3$, so $\dfrac{dy}{dx}=-\dfrac1{2\sqrt{9-x^2}}$. …