Question

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

Answer 1

`(e^4-1)/(1+4e^4)`

Explanation:

Apply the product rule:

`dy/dx*e^x+d/dx(e^x)*y-1*e^(2y)-d/dx(e^(2y))*x=d/dx(C)`

To find the derivative of `e^(2y)`, the chain rule will be necessary.

`d/dx(e^(2y))=e^(2y) * d/dx(2y)`
`=>e^(2y)* 2 * dy/dx`

Plug this back into the differentiated equation and simplify the other derivatives:

`dy/dx e^x+ye^x-e^(2y)-dy/dx 2xe^(2y)=0`

Solve for `dy/dx`.

`dy/dx e^x-dy/dx2xe^(2y)=e^(2y)-ye^x`

Continued simplification yields:

`dy/dx=(e^(2y)-ye^x)/(e^x-2xe^(2y))`

To find the slope of the tangent line, plug in the `x` and `y` values `(-2,1)`.

`dy/dx=(e^(2xx1)-(1)e^-2)/(e^-2-2(-2)e^(2xx1))`

Simplify:

`dy/dx=(e^4-1)/(1+4e^4)approx0.2443`

The slope could be written in either form—I prefer the more exact version, but knowing the decimal equivalence can be useful for checking your answer or for a continued application.