What is the derivative of #sin(x^2y^2)#?

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Answer 1 · 2015-03-24T20:57:41

Answer 1
If you want the partial derivatives of #f(x,y)=sin(x^2y^2)#, they are:

#f_x(x,y)=2xy^2cos(x^2y^2)# and

#f_y(x,y)=2x^2ycos(x^2y^2)#.

Answer 2
If we are considering #y# to be a function of #x# and looking for #d/(dx)(sin(x^2y^2))#, the the answer is:

#d/(dx)(sin(x^2y^2))=[2xy^2+2x^2y (dy)/(dx)]cos(x^2y^2)#

Find this using implicit differentiation (the chain rule) and the product rule.

#d/(dx)(sin(x^2y^2))=[cos(x^2y^2)]*d/(dx)(x^2y^2)#

#==[cos(x^2y^2)]*[2xy^2+x^2 2y (dy)/(dx)]#

#=[2xy^2+2x^2y (dy)/(dx)]cos(x^2y^2)#