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Question: If \({{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}}\), then \(\dfrac{dy}{dx}\) A....

If x23+y23=a23{{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}}, then dydx\dfrac{dy}{dx}
A. (y3)13{{\left( \dfrac{y}{3} \right)}^{\dfrac{1}{3}}}
B. (−yx)13{{\left( -\dfrac{y}{x} \right)}^{\dfrac{1}{3}}}
C. (xy)13{{\left( \dfrac{x}{y} \right)}^{\dfrac{1}{3}}}
D. (−xy)13{{\left( -\dfrac{x}{y} \right)}^{\dfrac{1}{3}}}

Explanation

Solution

We try to form the indices formula for the value 23\dfrac{2}{3}. We take the indices form of x23,y23,a23{{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 ddx(xn)=nxn−1\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 dydx\dfrac{dy}{dx} form of x23+y23=a23{{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 x23,y23,a23{{x}^{\dfrac{2}{3}}},{{y}^{\dfrac{2}{3}}},{{a}^{\dfrac{2}{3}}}.
We find the individual differentiations. So,
ddx(x23)+ddx(y23)=ddx(a23)\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 ddx(xn)=nxn−1\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}.

We find the derivatives of

\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]$$.