Question

How do you differentiate `z=A/y^10+Be^y`?

Answer 1

`dz/dy=-A/(10n^11)+Be^y`

Explanation:

Assuming `z` is a function and `A` and `B` are constants, we see that:

`dz/dy=d/dy(A/y^10)+d/dy(Be^y)`

Pulling out the constants:

`dz/dy=Ad/dy(y^-10)+Bd/dy(e^y)`

Now using `d/dy(y^n)=ny^(n-1)` and `d/dy(e^y)=e^y`, we see that:

`dz/dy=A(-10n^(-10-1))+B(e^y)`

`dz/dy=-(10A)/(n^11)+Be^y`