Question

Find the differentiation of given algebraic equation $\dfrac{dy}{dx}$if ${{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}}$

Answer 1

Hint: We know that on differentiation of ${x}^{n}$ power of x becomes less than 1 and the derivative of constant is to be zero. So,here we use the derivative formulas of ${x}^{n}$ and constant.

Complete step-by-step solution -
We are given ${{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}}$
The equation can be re-written as ${{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}-{{a}^{\dfrac{2}{3}}}=0$
Clearly , it is an implicit function . Hence , we can represent the function as $f(x,y)\equiv {{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}-{{a}^{\dfrac{2}{3}}}=0$
Now , we will differentiate the function $f(x,y)$with respect to $x$.
$f(x,y)$ is an implicit function. So , to differentiate it, we will apply the chain rule of differentiation . According to the chain rule of differentiation , if $R(x,y)$ is an implicit function, then the derivative of $R(x,y)$ with respect to $x$ is given by
$\dfrac{dR}{dx}=\dfrac{\partial R}{\partial x}+\dfrac{\partial R}{\partial y}\times \dfrac{dy}{dx}$
So, on differentiating both sides with respect to $x$, we get ,
$\dfrac{2}{3}.{{x}^{\left( \dfrac{2}{3}-1 \right)}}+\dfrac{2}{3}.{{y}^{\left( \dfrac{2}{3}-1 \right)}}.\dfrac{dy}{dx}=0$
$\dfrac{2}{3}{{x}^{\left( -\dfrac{1}{3} \right)}}+\dfrac{2}{3}{{y}^{\left( -\dfrac{1}{3} \right)}}.\dfrac{dy}{dx}=0$
Now , we need to find the value of $\dfrac{dy}{dx}$ . So , we will keep the terms with $\dfrac{dy}{dx}$ on one side and all other terms on the other side.
On keeping terms with $\dfrac{dy}{dx}$ on one side and all other terms on the other side , we get
$\dfrac{2}{3}{{y}^{\left( -\dfrac{1}{3} \right)}}.\dfrac{dy}{dx}=-\dfrac{2}{3}{{x}^{\left( -\dfrac{1}{3} \right)}}$
Now, we will shift $\dfrac{2}{3}{{y}^{\left( -\dfrac{1}{3} \right)}}$ from the left-hand side to the right-hand side.
On shifting $\dfrac{2}{3}{{y}^{\left( -\dfrac{1}{3} \right)}}$ from the left-hand side to the right-hand side , we get ,
$\dfrac{dy}{dx}=\dfrac{-{{x}^{\dfrac{-1}{3}}}}{{{y}^{\dfrac{-1}{3}}}}$
$\Rightarrow \dfrac{dy}{dx}=-{{\left( \dfrac{x}{y} \right)}^{\dfrac{-1}{3}}}$
Now, we will remove the minus sign from the exponent.
$\Rightarrow \dfrac{dy}{dx}=-{{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{3}}}$
Hence , the value of the derivative of the function ${{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}}$ is given by $\dfrac{dy}{dx}=-{{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{3}}}$.

Note: In the question , ${{a}^{\dfrac{2}{3}}}$ is an arbitrary constant. So , on differentiating it with respect to $x$, we get $\dfrac{d}{dx}\left( {{a}^{\dfrac{2}{3}}} \right)=0$
Some students consider it as a function of $x$ and write $\dfrac{d}{dx}\left( {{a}^{\dfrac{2}{3}}} \right)=\dfrac{2}{3}{{a}^{\dfrac{-1}{3}}}\dfrac{da}{dx}$
Such mistakes should be avoided . Such mistakes will create confusion while solving the problem as the student will not be able to find the value of $\dfrac{da}{dx}$.