How do you implicitly differentiate #2=e^(x-y)-xcosy #?

Question

1448 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-24T14:09:44

#(dy)/(dx) = (2+(x-1)cosy)/(2+x(cosy -siny))#

We have:

#e^(x-y) -xcosy = 2#

Differentiate both sides of the equation with respect to #x# keeping in mind that:

#d/(dx) f(y(x)) = f'(y(x)) * y'(x)#

so that:

#d/(dx) (e^(x-y) -xcosy) = 0#

#(1-y')e^(x-y) -cosy +xy'siny = 0#

Solving for #y'#:

#y'(e^(x-y) -xsiny) = e^(x-y) - cosy#

#y' = (e^(x-y) - cosy)/(e^(x-y) -xsiny)#

We can now simplify the expression substituting:

#e^(x-y) = 2+xcosy#

and we have:

#y' = (2+xcosy - cosy)/(2+xcosy -xsiny)#

#y' = (2+(x-1)cosy)/(2+x(cosy -siny))#