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

Concept: Chain Rule — $y = e^{\sin^{-1}x}$ is $e^{u}$ with $u=\sin^{-1}x$, so differentiate the outer exponential and multiply by the derivative of the inner inverse-sine.

1. $\dfrac{dy}{du} = e^{u} = e^{\sin^{-1}x}$.
2. $\dfrac{du}{dx} = \dfrac{1}{\sqrt{1-x^2}}$, for $|x|<1$.
3. Multiply: $\dfrac{dy}{dx} = e^{\sin^{-1}x}\cdot\dfrac{1}{\sqrt{1-x^2}}$.

The function $y = e^{\sin^{-1} x}$ is a composition: an outer exponential $e^{u}$ wrapped around an inner inverse-sine $u = \sin^{-1} x$. Whenever one function sits inside another like this, the chain rule is the tool to differentiate it — differentiate the outer function with respect to the inner, then multiply by the derivative of the inner function.

Step 1 — Identify the composition.

Let $u = \sin^{-1} x$, so $y = e^{u}$.

Step 2 — Differentiate the outer function.

$\dfrac{dy}{du} = e^{u} = e^{\sin^{-1} x}$ (the exponential is its own derivative).

Step 3 — Differentiate the inner function.

$\dfrac{du}{dx} = \dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}}$, defined for $|x| < 1$.

Step 4 — Multiply (chain rule).

Method: Chain Rule with an Inverse Trigonometric Inner Function

Use this whenever the function is an exponential (or any standard outer function) applied to an inverse trig function of $x$.

Steps

Step 1: Identify the outer and inner functions

Here the outer function is $e^{u}$ and the inner function is $u = \sin^{-1}x$.

Step 2: Differentiate the outer function, keeping the inner function intact

Step 3: Differentiate the inner inverse trig function using its standard derivative

Memorise (or quickly re-derive by implicit differentiation of $\sin y = x$) the standard derivatives of $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$ — they appear constantly as inner functions.

Step 4: Multiply the two results (chain rule) and state the domain

Always carry forward any domain restriction from the inner function's derivative (here $|x|<1$) into the final answer.

Common Mistakes

Mistake 1: Forgetting to multiply by the derivative of the inner function

Why it's wrong: writing $\dfrac{dy}{dx} = e^{\sin^{-1}x}$ and stopping ignores the chain rule entirely. Correct approach: always multiply the outer derivative by the inner function's derivative.

Mistake 2: Misremembering the derivative of $\sin^{-1}x$

Why it's wrong: some students write $\dfrac{d}{dx}\sin^{-1}x = \dfrac{-1}{\sqrt{1-x^2}}$ (that's actually the derivative of $\cos^{-1}x$), swapping the sign between the two. Correct approach: keep a clear anchor — $\sin^{-1}x$ increases, so its derivative is positive; $\cos^{-1}x$ decreases, so its derivative is negative.

Mistake 3: Dropping the domain restriction $|x|<1$

Why it's wrong: the derivative formula for $\sin^{-1}x$ is undefined at $x=\pm1$, so presenting the final derivative without noting where it's valid is an incomplete answer. Correct approach: state the domain restriction alongside the final derivative.