Implicit Derivation: Show That y' Satisfies Equation x sin(xy)+2x²=0

[superdave]

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.

Help needed please.

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]

$$\frac{d}{dx} (x \sin xy + 2x^2) = \sin xy + x \frac{d}{dx} \sin xy + 4x$$
$$= \sin xy + x (\cos xy \times \frac{d}{dx} xy) + 4x$$
$$= \sin xy + x (\cos xy \times (x \frac{dy}{dx} + y)) + 4x$$
$$= \sin xy + x \cos xy (xy' + y) + 4x$$
$$= \sin xy + xy \cos xy + y'x^2 \cos xy + 4x$$

 

[HallsofIvy]

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.

Help needed please.

I found the derivative of the first equation is:

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

In the first place x sin(xy)+ 2x² is not even an equation so it does not define y implicitely. I assume what you give was actually equal to some constant. In the second place, you did not find "the derivative of the first equation" because an equation does not have a derivative! What you did was differentiate the left hand side of your equation, treating y as if it were a constant. You cannot do that because y is not a constant, it is itself a function of x. You titled this "implicit differentiation" so you must have some idea what that is: use the chain rule. For example, the derivative of y² with respect to x is the derivative of y² with respect to y times the derivative of y with respect to x:
$$\frac{dy^2}{dx}= \frac{dy^2}{dy}\frac{dy}{dx}= 2y\frac{dy}{dx}$$
or just 2y y'. What is the derivative of sin(xy) with respect to x, remembering that y is an unknown function of y?