Question

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

Answer 1

Hint: 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${\left( {{x^2} + {y^2}} \right)^2}$, whereas product rule is used when differentiating two functions multiplied together, like $xy$.

Complete step-by-step answer:
On differentiating both sides with respect to $x$, we get
$ \Rightarrow 2\left( {{x^2} + {y^2}} \right)\dfrac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = x\dfrac{{dy}}{{dx}} + y$
$ \Rightarrow \left( {2{x^2} + 2{y^2}} \right)\left( {2x + 2y\dfrac{{dy}}{{dx}}} \right) = x\dfrac{{dy}}{{dx}} + 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$
$ \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$
$ \Rightarrow \dfrac{{dy}}{{dx}}\left[ {4{x^2}y + 4{y^3} - x} \right] = y - 4{x^3} - 4x{y^2}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y - 4{x^3} - 4x{y^2}}}{{4{x^2}y + 4{y^3} - 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., $\dfrac{d}{{dx}}\left[ {f\left( x \right) \cdot g\left( x \right)} \right] = f\left( x \right) \cdot g'\left( x \right) + g\left( x \right) \cdot f'\left( x \right)$.