Question

How do you implicitly differentiate `csc(x^2/y^2)=e^(xy)`?

Answer 1

`dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/(xy(2xcsc(x^2/y^2)cot(x^2/y^2)-ye^(xy))`

Explanation:

Implicit differentiation is no different from explicit differentiation. Just remember that differentiating a function of `y` causes the chain rule to be in effect. Recall also that `d/dxcscx=-cscxcotx` and `d/dxe^x=e^x`.

`csc(x^2/y^2)=e^(xy)`

`d/dxcsc(x^2/y^2)=d/dxe^(xy)`

`-csc(x^2/y^2)cot(x^2/y^2)*d/dx(x^2y^-2)=e^(xy)*d/dx(xy)`

Use the product rule to find these derivatives. Recall that while `d/dxx^2=2x`, `d/dxy^2=2ydy/dx`.

`-csc(x^2/y^2)cot(x^2/y^2)(2xy^-2-2x^2y^-1dy/dx)=e^(xy)(y+xdy/dx)`

Expanding and rearranging to group `dy/dx` terms:

`((2x^2csc(x^2/y^2)cot(x^2/y^2))/y-xe^(xy))dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2))/y^2+ye^(xy)`

Common denominators:

`((2x^2csc(x^2/y^2)cot(x^2/y^2)-xye^(xy))/y)dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/y^2`

Solving for `dy/dx`:

`dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/y^2*y/(2x^2csc(x^2/y^2)cot(x^2/y^2)-xye^(xy))`

`dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/(xy(2xcsc(x^2/y^2)cot(x^2/y^2)-ye^(xy))`