How do you implicitly differentiate # ysinx^2 = xe^-ysin(e^y)#?

Question

How do you implicitly differentiate #  ysinx^2 = xe^-ysin(e^y)#?

Evelyn Agard · Accepted Answer

#dy/dx.sinx^2 + 2xycosx^2 = e^-y sine^y + x( -e^-y. dy/dx.sine^y + dy/dx cos e^y) #

Adam Adkins · Answer

undefined To implicitly differentiate $ y\sin(x^2) = x e^{-y}\sin(e^y) $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Use the product rule and chain rule where necessary.
3. Solve for $ \frac{dy}{dx} $ after differentiation.

After differentiation, the result is:

$$ \frac{d}{dx}\left(y\sin(x^2)\right) = \frac{d}{dx}\left(x e^{-y}\sin(e^y)\right) $$

$$ y\cdot\frac{d}{dx}(\sin(x^2)) + \sin(x^2)\cdot\frac{dy}{dx} = e^{-y}\sin(e^y) + x\left(e^{-y}\cdot\frac{d}{dx}(\sin(e^y)) - \sin(e^y)\cdot\frac{dy}{dx}e^{-y}\right) $$

$$ y\cos(x^2)2x + \sin(x^2)\cdot\frac{dy}{dx} = e^{-y}\sin(e^y) + x\left(e^{-y}\cos(e^y)\cdot\frac{dy}{dx} - \sin(e^y)\cdot\frac{dy}{dx}e^{-y}\right) $$

$$ \frac{dy}{dx}\left(\sin(x^2) - x\sin(e^y)e^{-y}\right) = e^{-y}\sin(e^y) - 2xy\cos(x^2) - xe^{-y}\cos(e^y) $$

$$ \frac{dy}{dx} = \frac{e^{-y}\sin(e^y) - 2xy\cos(x^2) - xe^{-y}\cos(e^y)}{\sin(x^2) - x\sin(e^y)e^{-y}} $$