# 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

## "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)

(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)

## 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)

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

--Cori

## To find the derivative of f(x) = ln(x)(7x-2)^3, you can use the product rule, which states that the derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.

Step 1: Identify the functions and their derivatives:

- First function: ln(x)

- Derivative of the first function: d/dx [ln(x)]

Step 2: Find the derivative of ln(x):

The derivative of ln(x) can be found using the chain rule. The chain rule states that if you have a composite function, f(g(x)), then the derivative is given by f'(g(x)) * g'(x). In this case, f(u) = ln(u), and g(x) = x. So, applying the chain rule:

d/dx [ln(x)] = d/dx [ln(u)] * d/dx [u]

= 1/u * d/dx [u], where u = x

= 1/x

Step 3: Apply the product rule:

Using the product rule, the derivative of f(x) = ln(x)(7x-2)^3 is:

d/dx [ln(x)(7x-2)^3]

= (ln(x)) * d/dx [(7x-2)^3] + [(7x-2)^3] * d/dx [ln(x)]

Step 4: Find the derivatives of the second function:

To find the derivative of (7x-2)^3, you can use the power rule. The power rule states that if you have a function of the form f(x) = ax^n, then the derivative is given by d/dx [ax^n] = n * ax^(n-1). Applying the power rule:

d/dx [(7x-2)^3] = 3 * (7x-2)^2 * d/dx [7x-2]

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

= 21 * (7x-2)^2

Step 5: Substitute the derivatives back into the product rule equation:

d/dx [ln(x)(7x-2)^3]

= (ln(x)) * (21 * (7x-2)^2) + (7x-2)^3 * (1/x)

Simplifying the expression further, you can factor out (7x-2)^2 and x from the respective terms:

d/dx [ln(x)(7x-2)^3]

= 21 * ln(x) * (7x-2)^2 + (7x-2)^2 * (1/x)

So, the derivative of f(x) = ln(x)(7x-2)^3 is given by:

21 * ln(x) * (7x-2)^2 + (7x-2)^2 * (1/x)