Answer to Question #258535 in Calculus for Hamza

Question

Derivative of x^3+4xy^2+5y^3=1 at (2,-1)

Answer 1

Differentiate both sides with respect to $x$

$\dfrac{d}{dx}(x^3+4xy^2+5y^3)=\dfrac{d}{dx}(1)$

Use the Chain Rule

$3x^2+4y^2+8xy\dfrac{dy}{dx}+15y^2\dfrac{dy}{dx}=0$

Solve for $\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=-\dfrac{3x^2+4y^2}{8xy+15y^2}$

At $(2,-1)$

$\dfrac{dy}{dx}|_{(2,-1)}=-\dfrac{3(2)^2+4(-1)^2}{8(2)(-1)+15(-1)^2}=16$

$\dfrac{dy}{dx}|_{(2,-1)}=16$