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)
derivative of h(y)=log (y²+4y)
3 answers
sorry, but
(1 / (y² + 4y)) * (2y + 4) = 2(y+2)/(y²+4y)
(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)
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)