Question

What is the slope of the tangent line of `(y/x)e^(x/y-x^2)= C`, where C is an arbitrary constant, at `(1,2)`?

Answer 1

10

Explanation:

Taking the logarithm of both sides gives us

`ln y -ln x +x/y-x^2 = ln C`

Differentiating with respect to `x` yields

`1/y dy/dx -1/x +1/y-x/y^2dy/dx-2x = 0 implies`

`(1/y-x/y^2)dy/dx =2x+1/x-1/y implies`

`x(y-x)dy/dx = (2x^2+1)y^2-xy`

At the point `(1,2)` we have

`1 times (2-1)times dy/dx = (2 times1^2+1)times 2^2-1times 2 implies`
`dy/dx = 10`

Hence the slope of the tangent at `(1,2)` is 10

Note

I have assumed that tangent at `(1,2)` means tangent to the curve at `(1,2)`. Of course, this means that the constant `C` can not be arbitrary - but rather has to be `2/sqrt e`