Differentiate the following w.r.t. $x$. (i) $\cos ^{-1}(\sin x)$ (ii) $\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)$ (iii) $\sin ^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$

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

To differentiate the first expression, we use the identity $\sin x = \cos(\frac{\pi}{2} - x)$. This allows us to simplify the inverse cosine function. Since $\cos^{-1}(\cos \theta) = \theta$, the expression becomes $\frac{\pi}{2} - x$. Differentiating this with respect to $x$ gives us $-1$.

Step 2: Differentiate (ii) $\tan^{-1}\left(\frac{\sin x}{1+\cos x}\right)$

For the second expression, we use the half-angle trigonometric identities: $\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$ and $1+\cos x = 2\cos^2(\frac{x}{2})$. Substituting these into the expression inside the $\tan^{-1}$ function simplifies it to $\tan(\frac{x}{2})$. Then, $\tan^{-1}(\tan \theta) = \theta$, so the expression becomes $\frac{x}{2}$. Differentiating $\frac{x}{2}$ with respect to $x$ yields $\frac{1}{2}$.

Step 3: Differentiate (iii) $\sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$

For the third expression, we first rewrite the term inside the $\sin^{-1}$ function as $\frac{2 \cdot 2^x}{1+(2^x)^2}$. We then use a substitution: let $2^x = \tan \theta$. This transforms the expression into $\sin^{-1}(\frac{2\tan\theta}{1+\tan^2\theta})$, which simplifies to $\sin^{-1}(\sin 2\theta)$, or simply $2\theta$. Substituting back, we get $2\tan^{-1}(2^x)$. Finally, we differentiate $2\tan^{-1}(2^x)$ using the chain rule. The derivative of $\tan^{-1}(u)$ is $\frac{1}{1+u^2} \frac{du}{dx}$, and the derivative of $2^x$ is $2^x \log 2$. Combining these, we get the final result.

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