Question

What is the slope of the tangent line of `(x^2-y)e^y-y/x+x^3/y = C`, where C is an arbitrary constant, at `(2,2)`?

Answer 1

One form of the answer is `dy/dx=(8e^2+13)/(5-2e^2)`

Explanation:

Implicitly differentiate

`(2x-dy/dx)e^y+(x^2-y)e^ydy/dx-(dy/dxx-y)/x^2+(3x^2y-x^3dy/dx)/y^2=0`

Collect derivative terms

`-dy/dx[e^y(1-x^2+y)+1/x+x^3/y^2]+2xe^y+y/x^2+3x^2/y=0`

Solve for `dy/dx`

`dy/dx=(2xe^y+y/x^2+3x^2/y)/[e^y(1-x^2+y)+1/x+x^3/y^2]`

Substitute in the values `x=2` and `y=2`.

`dy/dx=(2*2e^2+2/2^2+3*2^2/2)/[e^2(1-2^2+2)+1/2+2^3/2^2]=(4e^2+1/2+6)/(-e^2+1/2+2)`

This will look a little simpler if we multiply the numerator and denominator by 2.

`dy/dx=(8e^2+13)/(5-2e^2)`