How do you find dy/dx for the function; cos(x^2+2Y)+xe^Y^2=1?

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Answer 1 · 2017-04-11T08:52:33

#dy/dx = (2x sin(x^2+2y)-e^(y^2))/( - 2sin(x^2+2y)+ 2xye^(y^2))#

Using the implicit function theorem:

#f(x,y) = cos(x^2+2y)+xe^(y^2) = 1#

#df = 0 = f_x dx + f_y dy implies dy/dx = - (f_x)/(f_y)#

So:

#f_x = -2x sin(x^2+2y)+e^(y^2)#

#f_y = - 2sin(x^2+2y)+ 2xye^(y^2)#

#dy/dx = - (f_x)/(f_y) = (2x sin(x^2+2y)-e^(y^2))/( - 2sin(x^2+2y)+ 2xye^(y^2))#