Q.Differentiate the following w.r.t. $x$:

Concept: Principal Value Domain — simplify each inverse-trig argument with an identity, watching the principal-value range at each step.

(i) $\cos^{-1}(\sin x)=\dfrac\pi2-x$ (principal range) $\Rightarrow$ derivative $-1$.

(ii) $\tan^{-1}\!\left(\dfrac{\sin x}{1+\cos x}\right)=\tan^{-1}\!\left(\tan\dfrac x2\right)=\dfrac x2$ $\Rightarrow$ derivative $\dfrac12$.

(iii) With $t=2^x=\tan\theta$, the argument is $\sin2\theta$. Since $\sin^{-1}(\sin2\theta)=2\theta$ only when $2\theta\le\dfrac\pi2$ (i.e. $x\le0$), and equals $\pi-2\theta$ when $2\theta>\dfrac\pi2$ (i.e. $x>0$), the derivative is piecewise:

Each part simplifies the inverse-trig argument with an identity first, then differentiates the simplified form — the efficient route, rather than differentiating the inverse-trig composition directly.

(i) $\cos^{-1}(\sin x)$

Using the complementary-angle identity $\sin x=\cos\left(\dfrac{\pi}{2}-x\right)$:

on the principal range. Differentiating,

(ii) $\tan^{-1}\!\left(\dfrac{\sin x}{1+\cos x}\right)$

Using half-angle identities $\sin x=2\sin\dfrac x2\cos\dfrac x2$ and $1+\cos x=2\cos^2\dfrac x2$:

So the function is $\tan^{-1}\!\left(\tan\dfrac x2\right) = \dfrac x2$ on the principal range, and

(iii) $\sin^{-1}\!\left(\dfrac{2^{x+1}}{1+4^x}\right)$

Put $t=2^x$ (so $t>0$ for every real $x$); then $4^x=t^2$ and $2^{x+1}=2t$, so the argument becomes $\dfrac{2t}{1+t^2}$.

This is where care is needed. Write $t=\tan\theta$ with $\theta\in\left(0,\dfrac{\pi}{2}\right)$ (possible since $t>0$). Then $\dfrac{2t}{1+t^2}=\sin2\theta$, so the expression is $\sin^{-1}(\sin2\theta)$ — but $\sin^{-1}(\sin y)=y$ only when $y$ lies in $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$; otherwise it equals $\pi-y$.

Since $\theta\in\left(0,\dfrac\pi2\right)$, $2\theta\in(0,\pi)$, which is not always inside $\sin^{-1}$'s principal range — so the simplification genuinely splits into two cases depending on the sign of $x$ (which controls whether $t=2^x\lessgtr1$, i.e. whether $\theta\lessgtr\dfrac\pi4$):

• If $x\le0$: $t=2^x\le1$, so $\theta\le\dfrac\pi4$ and $2\theta\le\dfrac\pi2$, which IS inside the principal range. Then

Differentiating (using $\dfrac{d}{dx}2^x=2^x\log 2$):

• If $x>0$: $t=2^x>1$, so $\theta>\dfrac\pi4$ and $2\theta>\dfrac\pi2$, OUTSIDE the principal range. Then

Differentiating:

So the derivative is genuinely piecewise, flipping sign at $x=0$ — a well-known trap in this exercise. (The function value itself is continuous at $x=0$, but the two one-sided derivatives there are $+\log 2$ and $-\log 2$, so $y$ is not differentiable at $x=0$.)

Method: Simplifying an Inverse Trig Composition Before Differentiating

This method applies whenever an inverse trig function wraps a trig expression that matches a standard identity — differentiating the raw, unsimplified expression directly (via chain and quotient rules) is far more error-prone than simplifying first.

Steps

Step 1: Look for a matching identity in the inner expression

Common patterns worth recognising: $\sin x=\cos\!\left(\tfrac{\pi}{2}-x\right)$; $\dfrac{\sin x}{1+\cos x}=\tan\tfrac{x}{2}$ (from half-angle identities); and $\dfrac{2t}{1+t^2}=\sin(2\theta)$ or $\dfrac{2t}{1-t^2}=\tan(2\theta)$ whenever $t=\tan\theta$ appears (here with $t$ replaced by an exponential like $2^x$).

Step 2: Substitute the identity to collapse the inverse function

Once the inner expression is rewritten as $\cos(\text{angle})$, $\tan(\text{angle})$, or $\sin(2\theta)$ in the matching form, the outer inverse function cancels the trig function directly (within its principal range), leaving a much simpler expression — often just a linear function of $x$ or a constant multiple of $\tan^{-1}$ of something simple.

Step 3: Differentiate the simplified expression

Now differentiate the reduced form using ordinary rules — this is typically a single application of the power/chain rule rather than a nested computation of chain-and-quotient rules on the original composite function.

Common Mistakes

Mistake 1: Differentiating the raw composite expression directly, without simplifying first

Why it's wrong: applying the chain rule to $\sin^{-1}$ or $\tan^{-1}$ of a full fraction (like $\dfrac{2^{x+1}}{1+4^x}$) directly requires a nested quotient-rule-inside-chain-rule computation that is slow and very easy to make an algebra slip in. Correct approach: always scan for a standard identity that collapses the inner expression first — the simplified form is almost always dramatically easier to differentiate.

Mistake 2: Applying an identity outside the range where it's actually valid

Why it's wrong: results like $\cos^{-1}(\sin x)=\tfrac{\pi}{2}-x$ or $\sin^{-1}\!\left(\dfrac{2t}{1+t^2}\right)=2\tan^{-1}t$ only hold on the identity's principal range — using them blindly for every $x$ without noting the restriction can silently produce a formula that fails outside that interval. Correct approach: state the identity's valid range alongside the simplification, even if the final differentiated answer looks the same within the intended domain.