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Question: If \({\left( {{x^2} + {y^2}} \right)^2} = xy\), find \(\dfrac{{dy}}{{dx}}\)....

If (x2+y2)2=xy{\left( {{x^2} + {y^2}} \right)^2} = xy, find dydx\dfrac{{dy}}{{dx}}.

Explanation

Solution

Chain rule and product rule are two really useful rules for differentiating functions. We use the chain rule when differentiating a ‘function of a function’, like(x2+y2)2{\left( {{x^2} + {y^2}} \right)^2}, whereas product rule is used when differentiating two functions multiplied together, like xyxy.

Complete step-by-step answer:
On differentiating both sides with respect to xx, we get
⇒2(x2+y2)ddx(x2+y2)=xdydx+y\Rightarrow 2\left( {{x^2} + {y^2}} \right)\dfrac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = x\dfrac{{dy}}{{dx}} + y
⇒(2x2+2y2)(2x+2ydydx)=xdydx+y\Rightarrow \left( {2{x^2} + 2{y^2}} \right)\left( {2x + 2y\dfrac{{dy}}{{dx}}} \right) = x\dfrac{{dy}}{{dx}} + y
⇒4x3+4x2ydydx+4xy2+4y3dydx=xdydx+y\Rightarrow 4{x^3} + 4{x^2}y\dfrac{{dy}}{{dx}} + 4x{y^2} + 4{y^3}\dfrac{{dy}}{{dx}} = x\dfrac{{dy}}{{dx}} + y
⇒4x3+4x2ydydx+4xy2+4y3dydx−xdydx−y=0\Rightarrow 4{x^3} + 4{x^2}y\dfrac{{dy}}{{dx}} + 4x{y^2} + 4{y^3}\dfrac{{dy}}{{dx}} - x\dfrac{{dy}}{{dx}} - y = 0
⇒dydx[4x2y+4y3−x]=y−4x3−4xy2\Rightarrow \dfrac{{dy}}{{dx}}\left[ {4{x^2}y + 4{y^3} - x} \right] = y - 4{x^3} - 4x{y^2}
⇒dydx=y−4x3−4xy24x2y+4y3−x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y - 4{x^3} - 4x{y^2}}}{{4{x^2}y + 4{y^3} - x}}
⇒dydx=y−4x(x2+y2)4y(x2+y2)−x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y - 4x\left( {{x^2} + {y^2}} \right)}}{{4y\left( {{x^2} + {y^2}} \right) - x}}

Note: The product rule states that differentiate a different function in the product each time and add the two terms together, i.e., ddx[f(x)⋅g(x)]=f(x)⋅g′(x)+g(x)⋅f′(x)\dfrac{d}{{dx}}\left[ {f\left( x \right) \cdot g\left( x \right)} \right] = f\left( x \right) \cdot g'\left( x \right) + g\left( x \right) \cdot f'\left( x \right).