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

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.


  • 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)= 1/x

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


    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.


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