# Implicit derivation

the equation x sin (xy) +2x² defines y implicitly as a function of x. assuming the derivative y' exists, show that it satisfies the equation y'x² cos (xy) +xy cos(xy)+sin (xy)+4x = 0.

I found the derivative of the first equation is:

sin xy + xy cos xy +4x. It's close to the answer, but not it.

James R
Homework Helper
Gold Member
$$\frac{d}{dx} (x \sin xy + 2x^2) = \sin xy + x \frac{d}{dx} \sin xy + 4x$$
$$= \sin xy + x (\cos xy \times \frac{d}{dx} xy) + 4x$$
$$= \sin xy + x (\cos xy \times (x \frac{dy}{dx} + y)) + 4x$$
$$= \sin xy + x \cos xy (xy' + y) + 4x$$
$$= \sin xy + xy \cos xy + y'x^2 \cos xy + 4x$$

HallsofIvy
Homework Helper
superdave said:
the equation x sin (xy) +2x² defines y implicitly as a function of x. assuming the derivative y' exists, show that it satisfies the equation y'x² cos (xy) +xy cos(xy)+sin (xy)+4x = 0.

$$\frac{dy^2}{dx}= \frac{dy^2}{dy}\frac{dy}{dx}= 2y\frac{dy}{dx}$$