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

Concept: Chain Rule — when a function is composed with another, differentiate the outer function and multiply by the derivative of the inner function.

1. $e^{-x}$ Outer: $e^u$, inner: $u = -x$. Derivative: $e^{-x} \cdot (-1) = -e^{-x}$.
2. $\sin(\log x)$, $x > 0$ Outer: $\sin u$, inner: $u = \log x$. Derivative: $\cos(\log x) \cdot \frac{1}{x} = \frac{\cos(\log x)}{x}$.
3. $\cos^{-1}(e^x)$ Outer: $\cos^{-1} u$, inner: $u = e^x$. Derivative: $\frac{-1}{\sqrt{1 - (e^x)^2}} \cdot e^x = -\frac{e^x}{\sqrt{1 - e^{2x}}}$.
4. $e^{\cos x}$ …

The Chain Rule is the backbone of differentiation when one function sits inside another. If you have $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$. Think of it as peeling an onion: differentiate the outer layer first, leaving the inner layer untouched, then multiply by the derivative of the inner layer. Each of these four problems is just that — a single chain, no nesting deeper than two functions.

Let’s work through them one by one.

1. Differentiate $e^{-x}$

   Here the outer function is $e^u$ (where $u = -x$), and the inner function is $u = -x$.

   The derivative of $e^u$ with respect to $u$ is $e^u$. So:

2. Differentiate $\sin(\log x)$, $x > 0$

   Outer: $\sin u$, inner: $u = \log x$.

   Derivative of $\sin u$ is $\cos u$, and derivative of $\log x$ is $\frac{1}{x}$. So:

3. Differentiate $\cos^{-1}(e^x)$

   Outer: $\cos^{-1} u$, inner: $u = e^x$.

   Recall the derivative of $\cos^{-1} u$ is $-\frac{1}{\sqrt{1-u^2}}$. So: …

Method: Applying the Chain Rule Across Different Elementary-Function Types

When a single question asks you to differentiate several composite functions built from different elementary functions (exponential, logarithmic, inverse trig), the approach is the same chain-rule pattern applied with a different "outer derivative formula" each time.

Steps

Step 1: For each part, isolate the outer function and note its standard derivative formula

Keep a short mental (or written) list of the standard derivatives you'll need:

Step 2: Identify the inner function $u$ for each part and differentiate it separately

E.g. for $\cos^{-1}(e^x)$, the inner function is $u = e^x$, so $\dfrac{du}{dx} = e^x$.

Step 3: Multiply the outer derivative (evaluated at the inner function) by the inner derivative

Step 4: Simplify each result and check the sign carefully …

Common Mistakes

Mistake 1: Dropping the negative sign on $e^{-x}$ or $e^{\cos x}\cdot(-\sin x)$

Why it's wrong: Whenever the inner function's derivative is itself negative (as with $-x$, or $-\sin x$ from $\cos x$), that negative sign must be carried through the multiplication — omitting it silently flips the sign of the whole answer. Correct approach: write out the inner derivative explicitly before multiplying, rather than multiplying "in your head" and guessing the sign at the end.

Mistake 2: Forgetting the $\frac{1}{x}$ factor from differentiating $\log x$ inside $\sin(\log x)$

Why it's wrong: The inner function here is $\log x$, whose derivative is $\frac{1}{x}$, not $1$ — leaving it out gives $\cos(\log x)$ as the final answer instead of the correct $\frac{\cos(\log x)}{x}$. Correct approach: always identify and differentiate the inner function completely, even when it's a "simple" function like $\log x$.

Mistake 3: Using the wrong sign in the derivative of $\cos^{-1}u$ …