NDA 2026 — Mathematics PYQ

Given the equation:

$$\sqrt[4]{y}=x+\sqrt{x^2+4}$$

1. Differentiate both sides with respect to $x$:

Using the chain rule on the left side:

$$\frac{d}{dx}(y^{1/4})=\frac{d}{dx}(x+\sqrt{x^2+4})$$

$$\frac{1}{4}{y}^{-3/4}\frac{dy}{dx}=1+\frac{1}{2\sqrt{x^2+4}}\cdot (2x)$$

2. Simplify the right side:

$$\frac{1}{4}{y}^{-3/4}\frac{dy}{dx}=1+\frac{x}{\sqrt{x^2+4}}$$

$$\frac{1}{4}{y}^{-3/4}\frac{dy}{dx}=\frac{\sqrt{x^2+4}+x}{\sqrt{x^2+4}}$$

3. Substitute the original equation:

Notice that the numerator $\sqrt{x^2+4}+x$ is exactly equal to $\sqrt[4]{y}$:

$$\frac{1}{4}{y}^{-3/4}\frac{dy}{dx}=\frac{\sqrt[4]{y}}{\sqrt{x^2+4}}$$

4. Isolate $\sqrt{x^2+4}\frac{dy}{dx}$:

Multiply both sides by $4\sqrt{x^2+4}\cdot y^{3/4}$:

$$\sqrt{x^2+4}\frac{dy}{dx}=4\cdot \sqrt[4]{y}\cdot y^{3/4}$$

Since $\sqrt[4]{y}\cdot y^{3/4}=y^{(1/4+3/4)}=y^1$:

$$\sqrt{x^2+4}\frac{dy}{dx}=4y$$

Final Answer:

The correct option is (d) $4y$.