Question

Can you please help?
dy/dx for (y^2 + 2e^(-xy) = 6)

Answer 1

∂/∂x or ∂/∂y?

Answer 2

um you're supposed to solve for dy/dx or y'...it's implicit differentiation I think

Answer 3

Implicit differentiation:\[\bf 2y \frac{ dy }{ dx }+2e^{-xy}(-y-x\frac{ dy }{ dx })=0\]\[\bf 2y \frac{ dy }{ dx }-2ye^{-xy}-2xe^{-xy}\frac{ dy }{ dx }=0\]Factor out dy/dx, rearrange, and solve for dy/dx:\[\bf \frac{dy}{dx}(2y-2xe^{-xy})-2ye^{-xy}=0\]\[\bf \frac{ dy }{ dx }(2y-2xe^{-xy})=2ye^{-xy} \implies \frac{dy}{dx}=\frac{ 2ye^{-xy} }{ 2y-2xe^{-xy} }=\frac{ ye^{-xy} }{ y-xe^{-xy} }\]

Answer 4

@luminate78

Answer 5

Thank you!!! :)