How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given $e^{sin y} + y = x^{2}$ $e^siny+y=x^2$ ?

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Answer 1 · 2016-08-30T19:23:48

$\frac{d y}{d x} = \frac{2 x}{e^{sin y} cos y + 1}$ $dy/dx=(2x)/(e^sinycosy+1)$

Differentiate both sides. Recall that the chain rule will be put into effect.

$\frac{d}{d x} (e^{sin y} + y = x^{2})$ $d/dx(e^siny+y=x^2)$

This gives us:

$e^{sin y} \frac{d}{d x} (sin y) + \frac{d y}{d x} = 2 x$ $e^sinyd/dx(siny)+dy/dx=2x$

Reapplying the chain rule:

$e^{sin y} cos y \frac{d y}{d x} + \frac{d y}{d x} = 2 x$ $e^sinycosydy/dx+dy/dx=2x$

Solving for $\frac{d y}{d x}$ $dy/dx$ :

$\frac{d y}{d x} (e^{sin y} cos y + 1) = 2 x$ $dy/dx(e^sinycosy+1)=2x$

Thus:

$\frac{d y}{d x} = \frac{2 x}{e^{sin y} cos y + 1}$ $dy/dx=(2x)/(e^sinycosy+1)$

How do you find (dy)/(dx)dydx(dy)/(dx) given e^siny+y=x^2esiny+y=x2e^siny+y=x^2?