How do you implicitly differentiate #2=e^ysinx-x^2y #?

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Answer 1 · 2017-01-23T00:03:19

#y'= (2xy-e^ycosx)/(2+x^2(y-1))#

Differentiate both sides of the equation with respect to #x#, keeping in mind that:

#d/dx(f(y(x)) = (df)/(dy)(dy)/(dx) = f'(y)y'#

#y'e^ysinx+ e^ycosx -2xy -x^2y' = 0#

Solve for #y'#:

#y'= (2xy-e^ycosx)/(e^ysinx-x^2)#

SUbstitute #e^siny =2+x^2y# from the original equation:

#y'= (2xy-e^ycosx)/(2+x^2y-x^2) = (2xy-e^ycosx)/(2+x^2(y-1))#