Differentiate the function.

Question

z(y) = (A) / (y^6) + Be^y

z'(y) = ?

Please show work and answer so i can get a better understanding of the problem

Answer 1

If A and B are constants then:

z(y) = (A) / (y^6) + Be^y

z(y) = Ay^-6 + Be^y

Using the power rule on the left portion we get:

[Ay^-6]' = (-6)Ay^-6-1 = -6Ay^-7 = -6A/y⁷

And using the exponential rule on the right side we get:

[Be^y]' = Be^y

So z'(y) = -6A/y⁷ + Be^y

or z'(y) = Be^y - 6A/y⁷