Solveeit Logo
Login

Question

Question: How do you determine \[\dfrac{{dy}}{{dx}}\] given \[{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\d...

How do you determine dydx\dfrac{{dy}}{{dx}} given x23+y23=a23{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}} ?

Explanation

Solution

Hint : The derivative is the rate of change of the quantity at some point. Now here in this question we consider the given function and we differentiate the given function with respect to x. by the standard differentiation formulas we differentiate. Hence, we can find the derivative of the function.
FORMULA USED:
ddx(xn)=n.xn−1ddx(x)\dfrac{d}{{dx}}({x^n}) = n.{x^{n - 1}}\dfrac{d}{{dx}}(x)

Complete step-by-step answer :
Here in this question, we can find the derivative by two methods.
Method 1: In this method consider the given function
x23+y23=a23{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}}
Apply the differentiation to the function
⇒ddx(x23)+ddx(y23)=ddx(a23)\Rightarrow \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 know that ddx(xn)=n.xn−1ddx(x)\dfrac{d}{{dx}}({x^n}) = n.{x^{n - 1}}\dfrac{d}{{dx}}(x) and the differentiation of a constant function is zero and applying this differentiation formula we have
⇒23x23−1dxdx+23y23−1dydx=0\Rightarrow \dfrac{2}{3}{x^{\dfrac{2}{3} - 1}}\dfrac{{dx}}{{dx}} + \dfrac{2}{3}{y^{\dfrac{2}{3} - 1}}\dfrac{{dy}}{{dx}} = 0
On simplifying we have
⇒23x2−33+23y2−33dydx=0\Rightarrow \dfrac{2}{3}{x^{\dfrac{{2 - 3}}{3}}} + \dfrac{2}{3}{y^{\dfrac{{2 - 3}}{3}}}\dfrac{{dy}}{{dx}} = 0
On further simplification we have
⇒23x−13+23y−13dydx=0\Rightarrow \dfrac{2}{3}{x^{\dfrac{{ - 1}}{3}}} + \dfrac{2}{3}{y^{\dfrac{{ - 1}}{3}}}\dfrac{{dy}}{{dx}} = 0
Take 23x−13\dfrac{2}{3}{x^{\dfrac{{ - 1}}{3}}} to the RHS of the equation and it is written as
⇒23y−13dydx=−23x−13\Rightarrow \dfrac{2}{3}{y^{\dfrac{{ - 1}}{3}}}\dfrac{{dy}}{{dx}} = - \dfrac{2}{3}{x^{\dfrac{{ - 1}}{3}}}
Cancel 23\dfrac{2}{3} on the both side of the equation we have
⇒y−13dydx=−x−13\Rightarrow {y^{\dfrac{{ - 1}}{3}}}\dfrac{{dy}}{{dx}} = - {x^{\dfrac{{ - 1}}{3}}}
Take y−13{y^{\dfrac{{ - 1}}{3}}} to RHS of the equation and it is written as
⇒dydx=−x−13y−13\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{{x^{\dfrac{{ - 1}}{3}}}}}{{{y^{\dfrac{{ - 1}}{3}}}}}
This is rewritten as
⇒dydx=−1x131y13\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\dfrac{1}{{{x^{\dfrac{1}{3}}}}}}}{{\dfrac{1}{{{y^{\dfrac{1}{3}}}}}}}
Taking reciprocal we have
⇒dydx=−y13x13\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{{y^{\dfrac{1}{3}}}}}{{{x^{\dfrac{1}{3}}}}}
Therefore dydx=−y13x13\dfrac{{dy}}{{dx}} = - \dfrac{{{y^{\dfrac{1}{3}}}}}{{{x^{\dfrac{1}{3}}}}}
Method 2: In this method consider the given equation
x23+y23=a23{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}}
This is rewritten as
⇒y23=a23−x23\Rightarrow {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}} - {x^{\dfrac{2}{3}}}
Applying the differentiation we have
⇒ddx(y23)=ddx(a23)−ddx(x23)\Rightarrow \dfrac{d}{{dx}}\left( {{y^{\dfrac{2}{3}}}} \right) = \dfrac{d}{{dx}}\left( {{a^{\dfrac{2}{3}}}} \right) - \dfrac{d}{{dx}}\left( {{x^{\dfrac{2}{3}}}} \right)
We know that ddx(xn)=n.xn−1ddx(x)\dfrac{d}{{dx}}({x^n}) = n.{x^{n - 1}}\dfrac{d}{{dx}}(x) and the differentiation of a constant function is zero and applying this differentiation formula we have
⇒23y23−1dydx=−23x23−1dxdx\Rightarrow \dfrac{2}{3}{y^{\dfrac{2}{3} - 1}}\dfrac{{dy}}{{dx}} = - \dfrac{2}{3}{x^{\dfrac{2}{3} - 1}}\dfrac{{dx}}{{dx}}
On simplifying we have
⇒23y2−33dydx=−23x2−33\Rightarrow \dfrac{2}{3}{y^{\dfrac{{2 - 3}}{3}}}\dfrac{{dy}}{{dx}} = - \dfrac{2}{3}{x^{\dfrac{{2 - 3}}{3}}}
On further simplification we have
⇒23y−13dydx=−23x−13\Rightarrow \dfrac{2}{3}{y^{\dfrac{{ - 1}}{3}}}\dfrac{{dy}}{{dx}} = - \dfrac{2}{3}{x^{\dfrac{{ - 1}}{3}}}
Cancel 23\dfrac{2}{3} on the both side of the equation we have
⇒y−13dydx=−x−13\Rightarrow {y^{\dfrac{{ - 1}}{3}}}\dfrac{{dy}}{{dx}} = - {x^{\dfrac{{ - 1}}{3}}}
Take y−13{y^{\dfrac{{ - 1}}{3}}} to RHS of the equation and it is written as
⇒dydx=−x−13y−13\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{{x^{\dfrac{{ - 1}}{3}}}}}{{{y^{\dfrac{{ - 1}}{3}}}}}
This is rewritten as
⇒dydx=−1x131y13\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\dfrac{1}{{{x^{\dfrac{1}{3}}}}}}}{{\dfrac{1}{{{y^{\dfrac{1}{3}}}}}}}
Taking reciprocal we have
⇒dydx=−y13x13\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{{y^{\dfrac{1}{3}}}}}{{{x^{\dfrac{1}{3}}}}}
Therefore dydx=−y13x13\dfrac{{dy}}{{dx}} = - \dfrac{{{y^{\dfrac{1}{3}}}}}{{{x^{\dfrac{1}{3}}}}}
Therefore, the derivative of x23+y23=a23{x^{\dfrac{2}{3}}} + {y^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3}}} is
−y13x13- \dfrac{{{y^{\dfrac{1}{3}}}}}{{{x^{\dfrac{1}{3}}}}}
Hence by the two methods we got the answer the same.
So, the correct answer is “ −y13x13 - \dfrac{{{y^{\dfrac{1}{3}}}}}{{{x^{\dfrac{1}{3}}}}} ”.

Note : To differentiate or to find the derivative of a function we use some standard differentiation formulas. The derivative is the rate of change of quantity, in this question we differentiate the given function with respect to x and find the derivative. For differentiation we must know the standard differentiation formulas