Question

How do you find the derivative `e^(x+y) + e^2x +e^2y`?

Answer 1

I assume that you want to find `d/dx(e^(x+y)+e^(2x)+e^(2y))`.

If I have misunderstood your question, please accept my apology.

`d/dx(e^(x+y)+e^(2x)+e^(2y)) = d/dx(e^(x+y))+d/dx(e^(2x))+d/dx(e^(2y))`.

For each term, we will need the chain rule applied to the exponential function:

`d/dx(e^u) = e^u (du)/dx`

`d/dx(e^(x+y)+e^(2x)+e^(2y)) = e^(x+y)d/dx(e^(x+y))+e^(2x)d/dx(2x)+e^(2y)d/dx(2y)`.

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

`= e^(x+y)+e^(x+y)dy/dx +2e^(2x)+2e^(2y)dy/dx`

So,
`d/dx(e^(x+y)+e^(2x)+e^(2y))=e^(x+y)+e^(x+y)dy/dx +2e^(2x)+2e^(2y)dy/dx`

If we had an equation, we could solve for `dy/dx`, but as it is, all we can do is rewrite using algebra:

`d/dx(e^(x+y)+e^(2x)+e^(2y))=[e^(x+y)+2e^(2x)]+[e^(x+y) +2e^(2y)]dy/dx`