derivative of h(y)=log (y²+4y)
sorry, but
(1 / (y² + 4y)) * (2y + 4) = 2(y+2)/(y²+4y)
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)
To find the derivative of h(y) = log(y^2 + 4y), we will use the chain rule and the derivative of the natural logarithm.
Step 1: Apply the chain rule:
The derivative of log(u) with respect to y is (1/u)*(du/dy).
In our case, u = y^2 + 4y, so du/dy = 2y + 4.
Step 2: Calculate the derivative using the chain rule:
h'(y) = (1/u)*(du/dy)
= (1/(y^2 + 4y))*(2y + 4)
So, the derivative of h(y) = log(y^2 + 4y) is h'(y) = (2y + 4)/(y^2 + 4y).
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)