How do you implicitly derive #e^cos y=x^3 arctan y#?

1 Answer
Mar 27, 2017

#(dy)/(dx)=-(3x^2arctany)/(sinye^cosy+x^3/(1+y^2))#

Explanation:

Implicit differentiation is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. #y=f(x)# - written explicitly as functions of #x#. However, some functions y are written implicitly as functions of #x#. So what we do is to treat #y# as #y=y(x)# and use chain rule. This means differentiating #y# w.r.t. #y#, but as we have to derive w.r.t. #x#, as per chain rule, we multiply it by #(dy)/(dx)#.

#e^cosy=x^3arctany#

then taking differential

#e^cosyxxd/(dy)cosyxx(dy)/(dx)=3x^2arctany+x^3xx1/(1+y^2)xx(dy)/(dx)#

or #-sinye^cosy(dy)/(dx)=3x^2arctany+x^3/(1+y^2)(dy)/(dx)#

or #(sinye^cosy+x^3/(1+y^2))(dy)/(dx)=-3x^2arctany#

and #(dy)/(dx)=-(3x^2arctany)/(sinye^cosy+x^3/(1+y^2))#