Question

How do you differentiate ${x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = 4$ ?

Answer 1

We can start with differentiating both the sides of the equation. We can use the Sum Rule, Power Rules and the Chain Rule here. After that we will solve as well as simplify all the terms to get the answer.

Complete step by step solution:
The given equation is: ${x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = 4$.
First, we have to start by differentiating on both the sides:
$\dfrac{d}{{dx}}\left( {{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}}} \right) = \dfrac{d}{{dx}}(4)$
When we apply the sum rule on the left side of the equation to the derivative $\left( {{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}}} \right)$ with respect to “x”, we get:
$\dfrac{d}{{dx}}\left( {{x^{\dfrac{2}{3}}}} \right) + \dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right)$

Now, we will solve $\dfrac{d}{{dx}}\left( {{x^{\dfrac{2}{3}}}} \right)$. We will differentiate by applying power rule that says:
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\,;\,where\,n = \dfrac{2}{3}$. Therefore, we will get:
$\dfrac{2}{3}\left( {{x^{\dfrac{2}{3} - 1}}} \right) + \dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right)$
$\Rightarrow\dfrac{2}{3}\left( {{x^{\dfrac{2}{3} - 1 \times \dfrac{3}{3}}}} \right) + \dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right)$ (Here, we are multiplying $- 1$ with $\dfrac{3}{3}$ to make $- 1$ a fraction)
$\Rightarrow \dfrac{2}{3}\left( {{x^{\dfrac{2}{3} + \dfrac{{ - 3}}{3}}}} \right) + \dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right)$
$\Rightarrow \dfrac{2}{3}\left( {{x^{\dfrac{{2 - 3}}{3}}}} \right) + \dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right)$

Now, for simplifying the numerator, we will multiply $- 1$ with $3$.
$\dfrac{2}{3}\left( {{x^{\dfrac{{2 - 3}}{3}}}} \right) + \dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right)$
$\Rightarrow\dfrac{2}{3}\left( {{x^{ - \dfrac{1}{3}}}} \right) + \dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right)$
Now, we will solve $\dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right)$
Now, we will use the Chain Rule here that says:
$\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right)\, = f'\left( {g\left( x \right)} \right)g'\left( x \right)\,\,where\,f\left( x \right) = {x^{\dfrac{2}{3}}};\,g\left( x \right) = y$
Here, we will change “y” as “u”:
$\dfrac{2}{3}{x^{ - \dfrac{1}{3}}} + \dfrac{d}{{du}}\left( {{u^{\dfrac{2}{3}}}} \right)\dfrac{d}{{dx}}\left( y \right)$

Now, I will replace the “y” in the place of “u”.
$\dfrac{2}{3}{x^{ - \dfrac{1}{3}}} + \dfrac{2}{3}\left( {{y^{\dfrac{2}{3} - 1}}} \right)\dfrac{d}{{dx}}\left( y \right)$
Now, we will again multiply $- 1$ with $3$, for simplifying the numerator:
$\Rightarrow \dfrac{2}{3}{x^{ - \dfrac{1}{3}}} + \dfrac{2}{3}\left( {{y^{\dfrac{2}{3} - 1 \times \dfrac{3}{3}}}} \right)\dfrac{d}{{dx}}\left( y \right)$
$\Rightarrow \dfrac{2}{3}{x^{ - \dfrac{1}{3}}} + \dfrac{2}{3}\left( {{y^{\dfrac{1}{3}}}} \right)\dfrac{d}{{dx}}\left( y \right)$
Now, we will try combine the terms that are together here:
$\Rightarrow\dfrac{2}{3}{x^{ - \dfrac{1}{3}}} + \dfrac{{2{y^{\dfrac{{ - 1}}{3}}}}}{3}\dfrac{d}{{dx}}\left( y \right)$
$\Rightarrow \dfrac{2}{3} \times \dfrac{1}{{{x^{\dfrac{1}{3}}}}} + \dfrac{2}{{3{y^{\dfrac{1}{3}}}}}\dfrac{d}{{dx}}\left( y \right)$

Now, we will try to simplify:
$\dfrac{2}{{3{x^{\dfrac{1}{3}}}}} + \dfrac{2}{{3{y^{\dfrac{1}{3}}}}}\dfrac{d}{{dx}}\left( y \right)$
We know that the derivative of $4$ with respect to “x” is $0$.
When we write the equation now:
$\dfrac{2}{{3{x^{\dfrac{1}{3}}}}} + \dfrac{2}{{3{y^{\dfrac{1}{3}}}}}y'\,\,\,where\,y' = \dfrac{d}{{dx}}(y)$
Now, we will solve for $y'$:
$\dfrac{2}{{3{x^{\dfrac{1}{3}}}}} + \dfrac{{2 \times y'}}{{3{y^{\dfrac{1}{3}}}}} = 0$
$\Rightarrow \dfrac{{2 \times y'}}{{3{y^{\dfrac{1}{3}}}}} = - \dfrac{2}{{3{x^{\dfrac{1}{3}}}}}$
$\Rightarrow 2y' = - \dfrac{2}{{3{x^{\dfrac{1}{3}}}}} \cdot \left( {3{y^{\dfrac{1}{3}}}} \right)$

Now, we will simplify $- \dfrac{2}{{3{x^{\dfrac{1}{3}}}}} \cdot \left( {3{y^{\dfrac{1}{3}}}} \right)$:
$\Rightarrow 2y' = - \dfrac{2}{{{x^{\dfrac{1}{3}}}}} \cdot \left( {{y^{\dfrac{1}{3}}}} \right)$
$\Rightarrow 2y' = - \dfrac{{2{y^{\dfrac{1}{3}}}}}{{{x^{\dfrac{1}{3}}}}}$
$\Rightarrow y' = - \dfrac{{{y^{\dfrac{1}{3}}}}}{{{x^{\dfrac{1}{3}}}}}$
We will now replace $\dfrac{{dy}}{{dx}}$in the place of $y'$:
$\therefore \dfrac{{dy}}{{dx}} = - \dfrac{{{y^{\dfrac{1}{3}}}}}{{{x^{\dfrac{1}{3}}}}}$

Note: This method is easy but it is very lengthy. There is another method which is easy as well as gets solved very quickly. The method is the implicit differentiation. In this method we will take “y” as the function of “x” and solve the equation.