Question

If ${{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}}$, then $\dfrac{dy}{dx}$
A. ${{\left( \dfrac{y}{3} \right)}^{\dfrac{1}{3}}}$
B. ${{\left( -\dfrac{y}{x} \right)}^{\dfrac{1}{3}}}$
C. ${{\left( \dfrac{x}{y} \right)}^{\dfrac{1}{3}}}$
D. ${{\left( -\dfrac{x}{y} \right)}^{\dfrac{1}{3}}}$

Answer 1

Hint: We try to form the indices formula for the value $\dfrac{2}{3}$. We take the indices form of ${{x}^{\dfrac{2}{3}}},{{y}^{\dfrac{2}{3}}},{{a}^{\dfrac{2}{3}}}$. We multiply the fraction with 3 to find the simplified form. The differentiation of constant gives 0. Then we use the formula of \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\] to find the derivatives.

Complete step by step answer:
We need to find the value of $\dfrac{dy}{dx}$ form of ${{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}}$. These are cube root forms.
The given value is the form of indices. We are trying to find the root value of ${{x}^{\dfrac{2}{3}}},{{y}^{\dfrac{2}{3}}},{{a}^{\dfrac{2}{3}}}$.
We find the individual differentiations. So,
$\dfrac{d}{dx}\left( {{x}^{\dfrac{2}{3}}} \right)+\dfrac{d}{dx}\left( {{y}^{\dfrac{2}{3}}} \right)=\dfrac{d}{dx}\left( {{a}^{\dfrac{2}{3}}} \right)$
We use the derivative formula of \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\].

We find the derivatives of
\[\dfrac{d}{dx}\left( {{x}^{\dfrac{2}{3}}} \right)=\dfrac{{{x}^{\dfrac{2}{3}-1}}}{\dfrac{2}{3}-1} \\
\Rightarrow \dfrac{d}{dx}\left( {{x}^{\dfrac{2}{3}}} \right)=-3{{x}^{-\dfrac{1}{3}}}\]
\[\Rightarrow \dfrac{d}{dx}\left( {{y}^{\dfrac{2}{3}}} \right)=\dfrac{{{y}^{\dfrac{2}{3}-1}}}{\dfrac{2}{3}-1}\dfrac{dy}{dx} \\
\Rightarrow \dfrac{d}{dx}\left( {{y}^{\dfrac{2}{3}}} \right)=-3{{y}^{-\dfrac{1}{3}}}\dfrac{dy}{dx}\],
$\Rightarrow \dfrac{d}{dx}\left( {{a}^{\dfrac{2}{3}}} \right)=0$.
For the differentiation of $\dfrac{d}{dx}\left( {{y}^{\dfrac{2}{3}}} \right)$, we used chain rule.
We now simplify the derivative equation.
\[-3{{x}^{-\dfrac{1}{3}}}-3{{y}^{-\dfrac{1}{3}}}\dfrac{dy}{dx}=0 \\
\Rightarrow 3{{y}^{-\dfrac{1}{3}}}\dfrac{dy}{dx}=-3{{x}^{-\dfrac{1}{3}}} \\
\Rightarrow \dfrac{dy}{dx}=-\dfrac{{{x}^{-\dfrac{1}{3}}}}{{{y}^{-\dfrac{1}{3}}}} \\
\Rightarrow \dfrac{dy}{dx}=-{{\left( \dfrac{x}{y} \right)}^{-\dfrac{1}{3}}} \\
\therefore \dfrac{dy}{dx}={{\left( -\dfrac{y}{x} \right)}^{\dfrac{1}{3}}} \]

Therefore, the correct option is B.

Note:The derivative form of \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\] is applicable for all values of $n\in \mathbb{R}\backslash \left\{ 0 \right\}$. We also could have used chain rule where we need remember that in the chain rule \[\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}\], we aren’t cancelling out the part \[d\left[ h\left( x \right) \right]\].