Question

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

Answer 1

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

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 `y` with respect to `x` will cause the chain rule to be in effect.

`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 `secy`):

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

Rewriting and grouping `dy/dx` terms:

`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)`