How do you implicitly differentiate #22=(y)/(x-e^y)#?

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Answer 1 · 2017-04-28T13:35:11

# color(red) (dy/dx = 22/(1+22e^y))#

Implicit differentiation is basically done in cases where #y# cannot be explicitly written as a function of #x#.

In this case,

#22=(y)/(x-e^y)#

#=> 22* (x-e^y) = y#

#=x-e^y=y/22 #

Differentiating both sides with respect to #x#,

#dx/dx - (de^y)/dx = 1/22*dy/dx#

Using chain rule to evaluate #(de^y)/dx#

#=> 1 - (de^y)/dy*dy/dx = 1/22dy/dx#

#=> 1 - e^ydy/dx=1/22dy/dx#

#=> (1/22+e^y)*dy/dx = 1#

#=> (1+22e^y)/22*dy/dx = 1#

#=> color(red) (dy/dx = 22/(1+22e^y))#