Question

If 16=2xy^2+x^2y find dy/dx(1,0)

Answer 1

We have the following function

$16=2xy^2+x^2y$

if we derive the function with respect to x we get

$\frac{d}{dx}\mleft(y\mright)=\frac{-2xy-2y^2}{x\left(x+4y\right)}$

if we evaluate the result function in point 1,0 we will obtain

$\frac{-2\cdot1\cdot0-2\cdot0}{1(1+4\cdot0)}=0$

Thus, dy/dx at point 1,0 is 0