Find the derivative.

y=e^8x/[e^(8x)+9]

y' =

The answer I have is 8e^8x/[(e^8x+9)^2] but it marked it incorrect..

okay here

set u=e^(8x)
v=(e^(8x)+9)

du/dx=8e^(8x)

dv/dx=8e^8x

dy/dx=[(e^8x+9)8e^8x-e^8x(8e^8x)]/(e^8x+9)^2

dy/dx=(8e^64x+72e^8x-8e^64x]/(e^8x+9)^2
=72e^8x/(e^8x+9)^2

check if i mad any erroe

I guess you did.

y' =
(8e^(8x))(e^(8x)+9)-(e^(8x))(8e^(8x))
-------------------------------
        (e^(8x)+9)^2

= 72e^(8x)/(e^(8x)+9)^2

You can use wolframalpha.com to confirm your results:

http://www.wolframalpha.com/input/?i=derivative+e%5E(8x)%2F%5Be%5E(8x)%2B9%5D

To find the derivative of the given function, we can use the quotient rule. The quotient rule states that if we have a function in the form of u/v, where u and v are both functions of x, then the derivative of the function is given by:

(f/g)' = (g * f' - f * g') / (g^2)

Let's apply this to the given function:

Given function: y = e^(8x) / (e^(8x) + 9)

To find the derivative, we need to find y':

Step 1: Find the derivative of the numerator (u'):

For u = e^(8x), the derivative is u' = (e^(8x))' = 8e^(8x)

Step 2: Find the derivative of the denominator (v'):

For v = e^(8x) + 9, the derivative is v' = (e^(8x) + 9)'

To find the derivative of e^(8x), we can use the chain rule. The chain rule states that if we have a function of the form f(g(x)), then the derivative is given by f'(g(x)) * g'(x). In this case, f(x) = e^x and g(x) = 8x.

So, applying the chain rule, the derivative of e^(8x) is (e^(8x))' = (e^x) * (8) = 8e^(8x)

Since the derivative of a constant (9 in this case) is zero, the derivative of v = e^(8x) + 9 is v' = 8e^(8x)

Step 3: Apply the quotient rule formula:

(f/g)' = (g * f' - f * g') / (g^2)

Substituting the values we found, we have:

y' = [(e^(8x) + 9) * 8e^(8x) - e^(8x) * 8e^(8x)] / (e^(8x) + 9)^2

Simplifying this expression further:

y' = [8e^(16x) + 72e^(8x) - 8e^(16x)] / (e^(8x) + 9)^2

y' = 72e^(8x) / (e^(8x) + 9)^2

So, the correct answer for y' is 72e^(8x) / (e^(8x) + 9)^2

Double-check the calculation steps to ensure accuracy.