Differentiate the following w.r.t. $x$ : (i) $e^{-x}$ (ii) $\sin (\log x), x>0$ (iii) $\cos ^{-1}\left(e^{x}\right)$ (iv) $e^{\cos x}$

Step 1: Differentiate $e^{-x}$

To differentiate $e^{-x}$, we use the chain rule. The derivative of $e^u$ with respect to $x$ is $e^u \cdot \frac{du}{dx}$. Here, $u = -x$. We differentiate $e^{-x}$ with respect to $-x$, and then multiply by the derivative of $-x$ with respect to $x$.

Step 2: Simplify the derivative of $e^{-x}$

The derivative of $-x$ with respect to $x$ is $-1$. Multiplying $e^{-x}$ by $-1$ gives us the final derivative.

Step 3: Differentiate $\sin (\log x)$

For $\sin (\log x)$, we again use the chain rule. The derivative of $\sin u$ is $\cos u \cdot \frac{du}{dx}$. Here, $u = \log x$. We differentiate $\sin (\log x)$ with respect to $\log x$, and then multiply by the derivative of $\log x$ with respect to $x$.

Step 4: Simplify the derivative of $\sin (\log x)$

The derivative of $\log x$ with respect to $x$ is $\frac{1}{x}$. Multiplying $\cos (\log x)$ by $\frac{1}{x}$ gives the final derivative.

Step 5: Differentiate $\cos^{-1}(e^x)$

We use the chain rule for $\cos^{-1}(e^x)$. The derivative of $\cos^{-1} u$ is $\frac{-1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$. Here, $u = e^x$. We differentiate $\cos^{-1}(e^x)$ with respect to $e^x$, and then multiply by the derivative of $e^x$ with respect to $x$.

Step 6: Simplify the derivative of $\cos^{-1}(e^x)$

The derivative of $e^x$ with respect to $x$ is $e^x$. Substituting this and simplifying gives the final derivative.

Step 7: Differentiate $e^{\cos x}$

For $e^{\cos x}$, we use the chain rule. The derivative of $e^u$ is $e^u \cdot \frac{du}{dx}$. Here, $u = \cos x$. We differentiate $e^{\cos x}$ with respect to $\cos x$, and then multiply by the derivative of $\cos x$ with respect to $x$.

Step 8: Simplify the derivative of $e^{\cos x}$

The derivative of $\cos x$ with respect to $x$ is $-\sin x$. Multiplying $e^{\cos x}$ by $-\sin x$ gives the final derivative.

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