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

1 Answer
Mar 5, 2017

#dy/dx = frac{(cos^2y)(e^(x-y))-cosy}{(cos^2y)(e^(x-y))+xsiny}#

Explanation:

#2=e^(x-y)-frac{x}{cosy}#

#0=(1-dy/dx)e^(x-y)-frac{cosy+xsiny(dy/dx)}{cos^2y}#

#0=e^(x-y)-e^(x-y)(dy/dx)-secy-(xsiny)/cos^2y (dy/dx)#

#0=dy/dx(-e^(x-y)-(xsiny)/cos^2y)+e^(x-y)-secy#

#dy/dx(e^(x-y)+(xsiny)/cos^2y)=e^(x-y)-secy#

#dy/dx = frac{e^(x-y)-secy}{(e^(x-y)+(xsiny)/cos^2y)}#

#dy/dx = frac{(cos^2y)(e^(x-y))-cosy}{(cos^2y)(e^(x-y))+xsiny}#

This fraction could be manipulated more to make terms cancel (eg multiplying by the conjugate of the denominator, etc.), but I have isolated #dy/dx#.