Question

Find $\frac{dy}{dx}$, if $x^3+8xy+y^3=64$.

Answer 1

$x^3+8xy+y^3=64$

Differentiating both sides w.r.t $x$ we get

$3x^2+8[x\frac{dy}{dx}+y]+3y^2\frac{dy}{dx}=0$

$\Rightarrow 8[x\frac{dy}{dx}+y]+3y^2\frac{dy}{dx}=-3x^2$

$\Rightarrow 8x\frac{dy}{dx}+8y+3y^2\frac{dy}{dx}=-3x^2$

$\Rightarrow 8x\frac{dy}{dx}+3y^2\frac{dy}{dx}=-3x^2-8y$

$\Rightarrow (8x+3y^2)\frac{dy}{dx}=-3x^2-8y$

$\Rightarrow \frac{dy}{dx}=\frac{-(3x^2+8y)}{(8x+3y^2)}$