# Help on differentiation

• Feb 10th 2008, 05:02 AM
Phatmat
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)
http://img151.imageshack.us/img151/6471/calca1av6.jpg

---------------------

Help on this entire question is needed
http://img151.imageshack.us/img151/5803/calca2or5.jpg

Thanks (Rofl),
Matt.
• Feb 10th 2008, 05:06 AM
galactus
2e is just a matter of the product rule.

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

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

Now, implement the rule.

$f(x)g'(x)+g(x)f'(x)$
• Feb 10th 2008, 05:10 AM
Phatmat
Though, to differentiate $ln f(x)$ is it:

f'(x)
----
f(x)
• Feb 10th 2008, 05:27 AM
galactus
Just use the chain rule.

$\frac{1}{2x}\cdot{2}$
• Feb 10th 2008, 09:34 AM
Jhevon
Quote:

Originally Posted by Phatmat
Help on this entire question is needed
http://img151.imageshack.us/img151/5803/calca2or5.jpg

Thanks (Rofl),
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

Quote:

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