Question

If $x^{2/3}+y^{2/3}=a^{2/3}$, then $\frac{dy}{dx}=$

• A. ${\left(\frac{y}{3}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
• B. ${\left(-\frac{y}{x}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
• C. ${\left(\frac{x}{y}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
• D. ${\left(-\frac{x}{y}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Answer 1

The correct option is B

${\left(-\frac{y}{x}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Explanation for the correct option:

Differentiation of given expression:

$x^{2/3}+y^{2/3}=a^{2/3}$

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

$$\begin{gathered} \frac{d{\left(x\right)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{dx}+\frac{d{\left(y\right)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{dx}=\frac{d{\left(a\right)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{dx} \\ \Rightarrow \frac{2{\left(x\right)}^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{3}+\frac{d{\left(y\right)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{dy}\times \frac{dy}{dx}=0\left[\because \frac{d}{dx}{x}^n=n{x}^{n-1}\right] \\ \Rightarrow \frac{2}{3}{\left(x\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+\frac{2}{3}{\left(y\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\times \frac{dy}{dx}=0 \\ \Rightarrow \frac{1}{x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}+\frac{1}{y^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\times \frac{dy}{dx}=0 \\ \Rightarrow \frac{1}{y^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\times \frac{dy}{dx}=-\frac{1}{x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \\ \Rightarrow \frac{dy}{dx}={\left(-\frac{y}{x}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \end{gathered}$$

Hence, option B is correct.