How do you differentiate #cos (y) -( x^2y^3) + 2y = pi#?

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Answer 1 · 2015-05-19T18:53:42

This is a bit of a vague question, as it can be differentiated in terms of either #x# or #y# or both #x and y#

I will assume for #x and y#

#d^2/(dxdy) cos(y) - (x^2y^3) + 2y = d^2/(dxdy) pi#

we then first differentiate over #y# and treat #x# as a constant

#d/dx -sin(y) - (3x^2y^2) + 2 = d/dx 0#

then we can differentiate over #x# and treat #y# as a constant

# -1 - (6xy^2) = 0#

#6xy^2 = -1#

#xy^2 = -1/6#