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

##### 1 Answer

Mar 19, 2018

#### Explanation:

#"differentiate "e^(x-y)cosy" using the "color(blue)"product rule"#

#0=e^(x-y)(-siny)dy/dx+cosye^(x-y)(1-dy/dx)#

#color(white)(0)=-e^(x-y)sinydy/dx+cosye^(x-y)-cosye^(x-y)dy/dx#

#rArrdy/dx(-e^(x-y)siny-cosye^(x-y))=-cosye^(x-y)#

#rArrdy/dx=(-cosye^(x-y))/(-e^(x-y)(siny+cosy))=(cosy)/(siny+cosy)#