Q.Differentiate the function $\sin^{-1}(x \sqrt{x}), 0 \leq x \leq 1$ with respect to $x$.

Concept: Chain Rule — simplify the argument, then differentiate $\sin^{-1}(u)$ using the chain rule.

Step 1: $x\sqrt x = x^{3/2}$, so $y = \sin^{-1}(x^{3/2})$.

Step 2: With $u = x^{3/2}$, $\dfrac{du}{dx} = \dfrac32x^{1/2}$ and $1-u^2 = 1-x^3$, so …

Step 1 — Simplify the argument

For $x\ge0$, $x\sqrt x=x\cdot x^{1/2}=x^{3/2}$. So

Step 2 — Apply the chain rule

With $u=x^{3/2}$, use $\dfrac{d}{dx}\sin^{-1}u=\dfrac{1}{\sqrt{1-u^2}}\cdot\dfrac{du}{dx}$:

So

Step 3 — Final form …

Method: Differentiating $\sin^{-1}(f(x))$ via the Chain Rule

For $y=\sin^{-1}(f(x))$, use the standard inverse-sine derivative $\dfrac{d}{dx}\sin^{-1}u=\dfrac{1}{\sqrt{1-u^2}}\cdot\dfrac{du}{dx}$ with $u=f(x)$ — first simplify $f(x)$ algebraically if possible, since a cleaner inner function makes the chain-rule step far easier.

Steps

Step 1: Simplify the argument of $\sin^{-1}$ algebraically if it can be written as a single power of $x$

Step 2: Apply the standard inverse-sine derivative with $u$ equal to that simplified argument

Step 3: Differentiate $u$ itself using the power rule, and substitute …

Common Mistakes

Mistake 1: Forgetting to simplify $x\sqrt{x}$ to $x^{3/2}$ before differentiating, and instead trying to differentiate the product $x\cdot\sqrt{x}$ inside the inverse sine directly.

Why it's wrong: while technically possible via the product rule on the inner function, it's far more error-prone than recognising $x\sqrt{x}=x^{3/2}$ first and differentiating a single power. Correct approach: always simplify the argument of an inverse trig function algebraically before applying the chain rule.

Mistake 2: Confusing "differentiate" with "find the principal value". …