Find the derivative $y'_x$ from the equation $y^3 + x^2 = xe^{y^2} - y\sin x$

Question

Find the derivative $y'_x$ (edit: $y'_x = {dy \over dx}$) from the equation $$y^3 + x^2 = xe^{y^2} - y\sin x$$

I generally don't understand how it should be done. Should I just implicitly differentiate it or there is another way? Should I include the logarithmic function somewhere since I have in the equation $e^{y^2}$?

Answer 1

Let $y(x)$ be the solution, then: $$ 0=\frac{\partial}{\partial x}(x^2+y(x)^3-x e^{y(x)^2}+y(x) \sin (x))= $$ $$ =3 y(x)^2 y'(x)-2 x e^{y(x)^2} y(x) y'(x)+\sin (x) y'(x)-e^{y(x)^2}+y(x) \cos(x)+2 x $$ With this we have: $$ y'(x)=\frac{-e^{y(x)^2}+y(x) \cos (x)+2 x}{-3 y(x)^2+2 x e^{y(x)^2} y(x)-\sin (x)}= $$ $$ =\frac{x^2-y(x)^3-y(x) \sin (x)+x y(x) \cos (x)}{x \left(2 x^2 y(x)+2 y(x)^4-3 y(x)^2+2 y(x)^2 \sin (x)-\sin (x)\right)} $$

Answer 2

Hint:

Deriving with respect to $x$ we have:

$$ \frac{d}{dx}\left(y^3+x^2-xe^{y^2}+y\sin x \right)=\frac{d}{dx}\left(y^3 \right)+\frac{d}{dx}\left(x^2 \right)-\frac{d}{dx}\left(xe^{y^2} \right)+\frac{d}{dx}\left(y \sin x \right)=0 $$ now use the product rule and the chain rule to evaluate the derivatives: $$ \frac{d}{dx}\left(y^3 \right)=2y^2 \frac{dy}{dx} $$ $$ \frac{d}{dx}\left(x^2 \right)=2x $$ $$ \frac{d}{dx}\left(xe^{y^2} \right)=e^{y^2}+2xye^{y^2} \frac{dy}{dx} $$

$$ \frac{d}{dx}\left(y \sin x \right)= \sin x\frac{dy}{dx}+y\cos x $$

now put all together and find $\frac{dy}{dx}$