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Question: Find \(\dfrac{{dy}}{{dx}}\) given \({y^3} + {x^3} = 3xy\)?...

Find dydx\dfrac{{dy}}{{dx}} given y3+x3=3xy{y^3} + {x^3} = 3xy?

Explanation

Solution

In this question, we have to find the derivative of the given quantity. Differentiate it with respect to x and use the product rule of the derivative on the right side. After that move the value which contains dydx\dfrac{{dy}}{{dx}} on one side and remaining on the other side. After that divide, both sides by the coefficient of dydx\dfrac{{dy}}{{dx}} and do simplification to get the desired result.

Complete step by step answer:
The given equation is y3+x3=3xy{y^3} + {x^3} = 3xy.
We have to find the value of dydx\dfrac{{dy}}{{dx}}.
So, differentiate the above equation with respect to xx what we have,
⇒ddx(y3+x3)=ddx(3xy)\Rightarrow \dfrac{d}{{dx}}\left( {{y^3} + {x^3}} \right) = \dfrac{d}{{dx}}\left( {3xy} \right)
Now apply the chain rule of differentiation on the right side, we have,
ddx(ab)=addx(b)+bddx(a)\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}\left( b \right) + b\dfrac{d}{{dx}}\left( a \right)
Apply the chain rule of differentiation on the right side, we have,
⇒ddx(y3+x3)=3xddx(y)+3yddx(x)\Rightarrow \dfrac{d}{{dx}}\left( {{y^3} + {x^3}} \right) = 3x\dfrac{d}{{dx}}\left( y \right) + 3y\dfrac{d}{{dx}}\left( x \right)
Differentiate the terms,
⇒3y2dydx+3x2=3xdydx+3y\Rightarrow 3{y^2}\dfrac{{dy}}{{dx}} + 3{x^2} = 3x\dfrac{{dy}}{{dx}} + 3y
Move dydx\dfrac{{dy}}{{dx}} terms on the left side and remaining on the right side,
⇒3y2dydx−3xdydx=3y−3x2\Rightarrow 3{y^2}\dfrac{{dy}}{{dx}} - 3x\dfrac{{dy}}{{dx}} = 3y - 3{x^2}
Take 3 commons on both sides,
⇒3(y2−x)dydx=3(y−x2)\Rightarrow 3\left( {{y^2} - x} \right)\dfrac{{dy}}{{dx}} = 3\left( {y - {x^2}} \right)
Cancel out common factors,
⇒(y2−x)dydx=(y−x2)\Rightarrow \left( {{y^2} - x} \right)\dfrac{{dy}}{{dx}} = \left( {y - {x^2}} \right)
Divide both sides by (y2−x)\left( {{y^2} - x} \right),
⇒dydx=(y−x2)(y2−x)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {y - {x^2}} \right)}}{{\left( {{y^2} - x} \right)}}
Hence, the value of dydx\dfrac{{dy}}{{dx}} for equation y3+x3=3xy{y^3} + {x^3} = 3xy is (y−x2)(y2−x)\dfrac{{\left( {y - {x^2}} \right)}}{{\left( {{y^2} - x} \right)}}.

Note: Differentiation can be defined as an independent variable value derivative that can be used to quantify features per unit change in an independent variable.
If the function f(x) undergoes an infinitesimal change of h in the vicinity of any point x, the function derivative is represented as
lim⁡h→0f(x+h)−f(x)h\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}