# How do you implicitly differentiate #9=e^y/sin^2x-e^x/cos^2y#?

##### 1 Answer

#### Explanation:

I'm going to use the fairly well-known trigonometric derivatives:

#d/dxcscx=-cscxcotx# #d/dxsecx=secxtanx#

First, I'll rewrite the function:

#9=e^ycsc^2x-e^xsec^2y#

When differentiating, we'll need to first use the product rule.

#0=(d/dxe^y)csc^2x+e^y(d/dxcsc^2x)-(d/dxe^x)sec^2y-e^x(d/dxsec^2y)#

Don't forget that differentiating a function of

#0=e^ydy/dxcsc^2x+e^y(2cscx)(d/dxcscx)-e^xsec^2y-e^x(2secy)(d/dxsecy)#

Applying the derivatives from before (and using the chain rule on the derivative of

#0=e^ydy/dxcsc^2x-2e^ycscx(cscxcotx)-e^xsec^2y-2e^xsecy(secytany)dy/dx#

Rewriting and grouping

#2e^ycsc^2xcotx+e^xsec^2y=dy/dx(e^ycsc^2x-2e^xsec^2ytany)#

Solve for the derivative:

#dy/dx=(2e^ycsc^2xcotx+e^xsec^2y)/(e^ycsc^2x-2e^xsec^2ytany)#