Q.Differentiate the following with respect to $x$: $\sin (\tan^{-1} e^{-x})$

Concept: Chain Rule — simplify $\sin(\tan^{-1}u)=\dfrac{u}{\sqrt{1+u^2}}$ first, then differentiate.

With $u=e^{-x}$: $y=\dfrac{e^{-x}}{\sqrt{1+e^{-2x}}} = e^{-x}(1+e^{-2x})^{-1/2}$.

Using the product rule with $p=e^{-x},\ q=(1+e^{-2x})^{-1/2}$: $p'=-e^{-x}$, $q'=e^{-2x}(1+e^{-2x})^{-3/2}$. …

Differentiating $y=\sin(\tan^{-1}e^{-x})$ directly (chain rule through sine, then arctan, then the exponential) is possible but algebraically messy. A cleaner route is to first simplify the composition $\sin(\tan^{-1}u)$ using a right triangle.

Step 1 — Simplify the inner composition.

For any real $u$, if $\theta=\tan^{-1}u$ then $\tan\theta=u$; picture a right triangle with opposite side $u$ and adjacent side $1$, so the hypotenuse is $\sqrt{1+u^2}$. Then $\sin\theta = \dfrac{u}{\sqrt{1+u^2}}$, i.e.

With $u=e^{-x}$,

Step 2 — Differentiate using the product rule.

Let $p=e^{-x}$ and $q=(1+e^{-2x})^{-1/2}$, so $y=pq$ and $y'=p'q+pq'$.

• $p' = -e^{-x}$.
• For $q$, apply the chain rule: $q' = -\dfrac12(1+e^{-2x})^{-3/2}\cdot\dfrac{d}{dx}(1+e^{-2x}) = -\dfrac12(1+e^{-2x})^{-3/2}\cdot(-2e^{-2x}) = e^{-2x}(1+e^{-2x})^{-3/2}$.

Step 3 — Combine. …

Method: Simplify a Trig-of-Inverse-Trig Composition with a Right-Triangle Identity Before Differentiating

Use this whenever you must differentiate an expression of the form $\sin(\tan^{-1}u)$, $\cos(\tan^{-1}u)$, $\tan(\sin^{-1}u)$, etc. — direct chain-rule differentiation is possible but produces messy $\sec^2$/$\sqrt{\ }$ expressions that are easy to mismanage.

Steps

Step 1: Build the reference right triangle for the inner inverse trig function

For $\theta = \tan^{-1}u$, picture a right triangle with opposite side $u$, adjacent side $1$, hypotenuse $\sqrt{1+u^2}$. This converts the inverse-trig angle into concrete side ratios.

Step 2: Read off the needed trig ratio from the triangle

This identity holds for all real $u$ and eliminates the inverse trig function completely.

Step 3: Substitute the given inner function for $u$ …

Common Mistakes

Mistake 1: Differentiating $\sin(\tan^{-1}e^{-x})$ directly, layer by layer, without simplifying first

Why it's wrong: this route works in principle but forces you to carry $\cos(\tan^{-1}e^{-x})$ and $\dfrac{1}{1+e^{-2x}}$ through several more algebra steps, multiplying the chance of a sign or simplification error. Correct approach: convert $\sin(\tan^{-1}u)$ to $\dfrac{u}{\sqrt{1+u^2}}$ first using the right-triangle identity, then differentiate the resulting algebraic expression.

Mistake 2: Dropping the negative sign when differentiating $e^{-x}$

Why it's wrong: $\dfrac{d}{dx}e^{-x} = -e^{-x}$, not $e^{-x}$ — missing this sign flips the sign of every subsequent term. Correct approach: always write the chain rule factor explicitly: $\dfrac{d}{dx}e^{-x} = e^{-x}\cdot(-1)$. …