Calculus
posted by Corin .
How would I go about finding the derivative of f(x)=ln(x)(7x2)^3
first=ln(x)
second= (7x2)^3
Would I do first times the derivative of the second + second times the derivative of the first?
If so, what is the derivative of lnx?
I'm confused.
Thanks for your help.
Cori

"Would I do first times the derivative of the second + second times the derivative of the first?"
That's right. But whenever you are not sure about such a rule you should derive it yourself from first principles. Otherwise you are just going to use a rule that you don't understand.
The derivative of ln(x) is 1/x. 
f(x)=lnx
f'(x)= 1/x
If I remember correctly your going to have to derize the second one again
(chain rule)
(7x2)^3
x= 7x2
dx= 7
(x)^3dx get the derivative of this
3(x)^31 dx + c
3(x)^2 dx + c
and plugging in the found values...
3(7x2)^2 ( 7)= 21(7x2)^2
(2nd part)
For the first part
it Should just be ln(x)= 1/x if I'm not incorrect (my text uses 2 functions instead of using ln so I'm not 100% sure)
so putting it together assuming my thinking is correct:
(1/x )(21(7x2)^2)
since the 2nd was already differentiated..
by the product rule if I remember correctly
f'(x)= (1/x)+(21(7x2)^2) 
I forgot a important part..the first part is to use the product rule
derivative of the first * second function + first*derivative of the second
then doing this again,correcting that error
f(x)= ln(x)(7x2)^3
product rule first then the chain rule for (7x2)^3
(1/x)(7x2)^3 + (lnx)3(7x2)^2
chain rule for the 2nd part
x = 7x2
dx = 7
~you could replace the internal equation 7x2 with x or not but
if you do
(1/x)(7x2)^3 + (lnx)3(x)^2dx
plug in the values of x and dx and
(1/x)(7x2)^3 + (lnx)3(7x2)^2(7)=
(1/x)(7x2)^3 + 21 (lnx)(7x2)^2
(I didn't go and simplify though) 
Thanks so much for all your help. I figured it out.
Cori