1. ## Help on differentiation

Hello to everyone, I have just signed up and hope that I can grow welcome on these forums.

I have a calculus assignment due in this week, and needed some help on a few questions, any help would be much appreciated.

Only need help on 2)e)

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Help on this entire question is needed

Thanks ,
Matt.

2. 2e is just a matter of the product rule.

$f(x)=e^{x}$

$g(x)=ln(2x)$

Now, implement the rule.

$f(x)g'(x)+g(x)f'(x)$

3. Though, to differentiate $ln f(x)$ is it:

f'(x)
----
f(x)

4. Just use the chain rule.

$\frac{1}{2x}\cdot{2}$

5. Originally Posted by Phatmat
Help on this entire question is needed

Thanks ,
Matt.
for 3(a), use the chain rule, the result follows immediately, you may want to write out some steps though, so it looks like you're doing something. remember... $\frac d{dx} h(k(x)) = h'(k(x)) \cdot k'(x)$

that is, you differentiate the entire function leaving the inside function intact (that is, you treat it as if it were a single variable), then multiply by the derivative of the inside function

it's awkward using h and k as the functions for this rule, but f was taken, so...

for 3(b): the product rule says: $\frac d{dx}h(x) \cdot k(x) = h'(x)k(x) + h(x)k'(x)$

just apply the product rule here (use part (a) to find the derivative of $[g(x)]^{-1}$). simplify, and you will get the quotient rule

Originally Posted by Phatmat
Though, to differentiate $ln f(x)$ is it:

f'(x)
----
f(x)
yes. as galactus said, we can see this by using the chain rule