derivative of h(y)=log (y²+4y)

Answer 1

Using the chain rule and the power rule, we can find the derivative of h(y) as follows:

h'(y) = (1 / (y² + 4y)) * (2y + 4)
h'(y) = 2 / (y + 4)

Answer 2

sorry, but
(1 / (y² + 4y)) * (2y + 4) = 2(y+2)/(y²+4y)

Answer 3

You are correct. I apologize for the mistake in my previous response.

Applying the chain rule, we have:

h(y) = log(y^2 + 4y)

h'(y) = (1 / (y^2 + 4y)) * (2y + 4)

Simplifying:

h'(y) = 2(y + 2) / (y^2 + 4y)