Question

Find dy/dx by implicit differentiation: x^2y^3-7xy^2=10

Answer 1

$x^2y^3-7xy^2=10$

Differentiate both sides wrt $x$:

$\dfrac{\mathrm d(x^2y^3-7xy^2)}{\mathrm dx}=\dfrac{\mathrm d(10)}{\mathrm dx}$

$\dfrac{\mathrm d(x^2y^3)}{\mathrm dx}-7\dfrac{\mathrm d(xy^2)}{\mathrm dx}=0$

By the product rule,

$\left(y^3\dfrac{\mathrm d(x^2)}{\mathrm dx}+x^2\dfrac{\mathrm d(y^3)}{\mathrm dx}\right)-7\left(y^2\dfrac{\mathrm d(x)}{\mathrm dx}+x\dfrac{\mathrm d(y^2)}{\mathrm dx}\right)=0$

By the power and chain rules,

$\left(y^3(2x)+x^2\left(3y^2\dfrac{\mathrm dy}{\mathrm dx}\right)\right)-7\left(y^2+x\left(2y\dfrac{\mathrm dy}{\mathrm dx}\right)\right)=0$

$\left(2xy^3+3x^2y^2\dfrac{\mathrm dy}{\mathrm dx}\right)-7\left(y^2+2xy\dfrac{\mathrm dy}{\mathrm dx}\right)=0$

$(3x^2y^2-14xy)\dfrac{\mathrm dy}{\mathrm dx}=7y^2-2xy^3$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{7y^2-2xy^3}{3x^2y^2-14xy}$

and if $y\neq0$,

$\implies\boxed{\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{7y-2xy^2}{3x^2y-14x}}$