# Calculus

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How would I go about finding the derivative of f(x)=ln(x)(7x-2)^3

first=ln(x)
second= (7x-2)^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

• Calculus -

"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.

• Calculus -

f(x)=lnx

f'(x)= 1/x

If I remember correctly your going to have to derize the second one again
(chain rule)

(7x-2)^3

x= 7x-2
dx= 7

(x)^3dx get the derivative of this

3(x)^3-1 dx + c
3(x)^2 dx + c

and plugging in the found values...

3(7x-2)^2 ( 7)= 21(7x-2)^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(7x-2)^2)

since the 2nd was already differentiated..
by the product rule if I remember correctly

f'(x)= (1/x)+(21(7x-2)^2)

• Calculus -

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)(7x-2)^3

product rule first then the chain rule for (7x-2)^3

(1/x)(7x-2)^3 + (lnx)3(7x-2)^2

chain rule for the 2nd part
x = 7x-2
dx = 7

~you could replace the internal equation 7x-2 with x or not but
if you do

(1/x)(7x-2)^3 + (lnx)3(x)^2dx

plug in the values of x and dx and

(1/x)(7x-2)^3 + (lnx)3(7x-2)^2(7)=

(1/x)(7x-2)^3 + 21 (lnx)(7x-2)^2

(I didn't go and simplify though)

• Calculus -

Thanks so much for all your help. I figured it out.

--Cori

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