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

We need to differentiate two distinct terms: $(\sin x)^x$ (a variable base and exponent — use logarithmic differentiation) and $\sin^{-1}\sqrt{x}$ (a standard inverse trig composition).

Step 1 — Differentiate $(\sin x)^x$

Let $y_1 = (\sin x)^x$. Take $\log y_1 = x \log(\sin x)$. Differentiate:

So

Step 2 — Differentiate $\sin^{-1}\sqrt{x}$

Let $y_2 = \sin^{-1}\sqrt{x}$. Using $\frac{d}{dx} \sin^{-1} u = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$ with $u = \sqrt{x}$: …

---

Why this approach works

The function has two very different pieces glued together by addition. The first term, $(\sin x)^x$, has the variable in both the base and the exponent — that’s a red flag for the standard power rule or exponential rule alone. The cleanest way is to take the natural logarithm first, differentiate implicitly, then solve for the derivative. This is called logarithmic differentiation.

The second term, $\sin^{-1}\sqrt{x}$, is a composition of the inverse sine function with a square root. That’s a straightforward chain rule job — no tricks.

We’ll handle each term separately, then add.

---

Step-by-step solution

1. Differentiate $(\sin x)^x$

Let $y_1 = (\sin x)^x$. Take the natural logarithm of both sides:

Now differentiate both sides with respect to $x$. On the left, by the chain rule:

On the right, use the product rule on $x \cdot \log(\sin x)$:

That simplifies to:

So we have:

Multiply through by $y_1$:

2. Differentiate $\sin^{-1}\sqrt{x}$

Let $y_2 = \sin^{-1}\sqrt{x}$. Recall the standard derivative:

Here $u = \sqrt{x} = x^{1/2}$, so $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$. …

Method: Differentiating a Sum of a Variable-Exponent Term and an Inverse Trigonometric Term

Use this method when an expression combines two structurally different pieces added together — one needing logarithmic differentiation (a variable base to a variable power), the other needing a standard inverse-trig derivative formula with the Chain Rule. Treat each piece completely independently.

Steps

Step 1: Split the sum and identify which rule each term needs.

Step 2: Log-differentiate the variable-base-and-exponent term.

Step 3: Differentiate the inverse-sine term using the standard formula plus the Chain Rule.

Recall $\dfrac{d}{dx}\sin^{-1}u = \dfrac{1}{\sqrt{1-u^2}}\cdot\dfrac{du}{dx}$. With $u=\sqrt{x}$, $\dfrac{du}{dx}=\dfrac{1}{2\sqrt{x}}$: …

Common Mistakes

Mistake 1: Missing the Chain Rule factor $\frac{1}{2\sqrt x}$ when differentiating $\sin^{-1}\sqrt x$.

Why it's wrong: writing $\frac{dy_2}{dx} = \frac{1}{\sqrt{1-x}}$ alone applies only the outer inverse-sine formula and forgets to multiply by the derivative of the inner function $\sqrt x$. Correct approach: always identify the inner function inside an inverse trig expression and multiply by its derivative, exactly as with any other composite function.

Mistake 2: Applying logarithmic differentiation to the whole sum instead of just the $(\sin x)^x$ term. …