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

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Answer 1 · 2016-08-30T19:23:48

#dy/dx=(2x)/(e^sinycosy+1)#

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

#d/dx(e^siny+y=x^2)#

This gives us:

#e^sinyd/dx(siny)+dy/dx=2x#

Reapplying the chain rule:

#e^sinycosydy/dx+dy/dx=2x#

Solving for #dy/dx#:

#dy/dx(e^sinycosy+1)=2x#

Thus:

#dy/dx=(2x)/(e^sinycosy+1)#