Differentiate the following w.r.t. x. (i) cos^-1(sin x) (ii) tan^-1(sin x / (1 + cos x)) (iii) sin^-1(2^$(x+1) / (1 + 4^x)$)

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

To differentiate $\cos^{-1}(\sin x)$, we use the chain rule. The derivative of $\cos^{-1}(u)$ is $\frac{-1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$. Here, $u = \sin x$.

Step 2: Simplify (i)

We know that $\frac{d}{dx}(\sin x) = \cos x$. Also, using the identity $\sin^2 x + \cos^2 x = 1$, we can replace $1 - \sin^2 x$ with $\cos^2 x$. The square root of $\cos^2 x$ is $|\cos x|$.

Step 3: Differentiate (ii) $\tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right)$

First, we simplify the argument of the $\tan^{-1}$ function using half-angle identities: $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1 + \cos x = 2 \cos^2(x/2)$. This simplifies the expression to $\tan(x/2)$.

Step 4: Differentiate simplified (ii)

Since $\tan^{-1}(\tan \theta) = \theta$, the expression simplifies to $x/2$. Now, we differentiate $x/2$ with respect to $x$.

Step 5: Differentiate (iii) $\sin^{-1}\left(\frac{2^{x+1}}{1 + 4^x}\right)$

Let's rewrite the argument of the $\sin^{-1}$ function. We can express $2^{x+1}$ as $2 \cdot 2^x$ and $4^x$ as $(2^x)^2$. This transformation prepares the expression for a substitution.

Step 6: Substitute and simplify (iii)

Let $2^x = \tan \theta$. Then the expression becomes $\frac{2 \tan \theta}{1 + \tan^2 \theta}$. This is a known trigonometric identity for $\sin(2\theta)$.

Step 7: Differentiate simplified (iii)

Since $\sin^{-1}(\sin(2\theta)) = 2\theta$, we substitute back $\theta = \tan^{-1}(2^x)$. Now we need to differentiate $2 \tan^{-1}(2^x)$ with respect to $x$.

Step 8: Final differentiation of (iii)

The derivative of $\tan^{-1}(u)$ is $\frac{1}{1 + u^2} \cdot \frac{du}{dx}$. Here $u = 2^x$, and $\frac{d}{dx}(2^x) = 2^x \ln 2$. Combining these, we get the final derivative.

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