⇒Derivative

Correct Option is (B)

$x^k+y^k=a^k$

Differentiating w.r.t. $x$, we get

$\begin{array}{rl}& {\mathrm{kx}}^{\mathrm{k}-1}+\mathrm{k}{y}^{\mathrm{k}-1}\frac{\mathrm{d}y}{\mathrm{d}x}=0\\ & \therefore \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{\mathrm{k}{x}^{k-1}}{\mathrm{k}{y}^{k-1}}\\ & \therefore \frac{\mathrm{d}y}{\mathrm{d}x}=-{\left(\frac{x}{y}\right)}^{\mathrm{k}-1}\\ & \frac{\mathrm{d}y}{\mathrm{d}x}+{\left(\frac{y}{x}\right)}^{1-k}=0\\ & \text{ But }\frac{\mathrm{d}y}{\mathrm{d}x}+{\left(\frac{y}{x}\right)}^{\frac{1}{3}}=0\text{... [Given]}\end{array}$

Comparing above equations, we get

$\begin{array}{rl}& 1-\mathrm{k}=\frac{1}{3}\\ & 1-\frac{1}{3}=\mathrm{k}\\ & \therefore \mathrm{k}=\frac{2}{3}\end{array}$

Correct Option is (C)

$y=\sqrt{e^{\sqrt{x}}}\Rightarrow y^2=e^{\sqrt{x}}$

Takig log on both sides,

$2\log y=\sqrt{x}\log e\Rightarrow 2\log y=\sqrt{x}$

Differentiating both sides w.r.t. $x$, we get

$\frac{2}{y}\frac{dy}{dx}=\frac{1}{2\sqrt{x}}\Rightarrow \frac{dy}{dx}=y\left[\frac{1}{4\sqrt{x}}\right]=\frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$