I have trouble finding the second derivative, my answer is a little off from the answer key.

f(x)=(log(base2)x)^5
f'(x)=[5(log(base2)x)^4]/xln2
f"(x)=[20(log(base2)x)³(xln2)-(ln2+(x/2))(5(log(base2)x)^4)]/[x²(ln2)²]

for the f"(x), is g(x)=xln2? It is a product so I think it should be dealt with differently, but I am unclear on that.

Thanks

in your f ''(x) you forgot to differentiate the log2x part , and in the second part when you differentiate xln2 you simply get ln2.

Remember ln2 is just a number so xln2 is (ln2)x , a simple term like 6x.

I had

f ''(x) = [20(log(base2)x)³(1/(xln2)(xln2)-(ln2))(5(log(base2)x)^4)]/[x²(ln2)²]

Now doesn't the xln2 at the top and bottom of the first section cancel?
check my arithmetic, too early in the morning

I have the same thing now, thanks! I thought I had to use product rule for xln2, but there's no variable in ln2, so it is accounted for a constant?

To find the second derivative of the function f(x) = (log(base2)x)^5, you correctly found the first derivative. Now, let's proceed to find the second derivative.

To understand this, let's break down the function into simpler parts. We have (log(base2)x)^5 = g(x)^5, where g(x) = log(base2)x.

Now, to calculate the second derivative, we can write f''(x) using the chain rule. The chain rule states that if we have a composite function u(v(x)), then its derivative is given by u'(v(x)) * v'(x). In our case, g(x) is our composite function, and we have f''(x) = [u(v(x))]' = u'(v(x)) * v'(x).

Let's calculate the second derivative step by step:

1. First, let's find g'(x), which is the derivative of g(x) = log(base2)x:
We can use the logarithmic differentiation rules:
g'(x) = (1/ln2) * (1/x) = 1/(x ln2)

2. Now, let's find u'(v(x)), which is the derivative of u(v(x)) = g(x)^5:
To differentiate u(v(x)) with respect to v(x), we treat g(x) as a constant. Thus, u'(v(x)) = 5(g(x))^4 * g'(x).

Plugging in the values we know:
u'(v(x)) = 5(log(base2)x)^4 * (1/(x ln2))

3. Finally, let's find v'(x), which is the derivative of v(x) = log(base2)x:
Applying the logarithmic differentiation rules again:
v'(x) = (1/ln2) * (1/x) = 1/(x ln2)

4. Now, we have all the values to find f''(x).
Using the chain rule, f''(x) = u'(v(x)) * v'(x):
f''(x) = 5(log(base2)x)^4 * (1/(x ln2)) * 1/(x ln2)

Simplifying the expression:
f''(x) = (5(log(base2)x)^4)/(x^2(ln2)^2)

Therefore, your expression for the second derivative f''(x) is correct.

In summary, when dealing with a product of functions, like g(x) = x ln2 in your case, we use the chain rule to find its derivative by multiplying the derivative of one factor with the other factor.