Question

How do you find the slope of the line tangent to `2sin^2x=3cosy` at `(pi/3,pi/3)`?

Answer 1

`(2/3)cos(x)/sin(cos^-1((2/3)sin(x)))=dy/dx`

Explanation:

Divide by 3 on both sides:

`(2/3)sin(x)=cos(y)`

Then use implicit differentiation. Note that the derivative of `sin(x)` is `cos(x)` and the derivative of `cos(x)` is `-sin(x)`. The `(2/3)` is unaffected

`(2/3)cos(x)=-sin(y) (dy/dx)`

Solve for `dy/dx`

`(2/3)cos(x)/sin(y)=dy/dx`

However, you can solve for y from the given equation and plug it in for y.

`(2/3)sin(x)=cos(y)`

`cos^-1((2/3)sin(x))=y`

`(2/3)cos(x)/sin(cos^-1((2/3)sin(x)))=dy/dx`