Question

how do i find the derivative of ln[(e ^(4^2)) (3x-1)^4 (x^3-1)^5].

Answer 1

This is a bit messy. I am assuming it's \(\ln[(e^{4^2} (3x-1)^4 (x^3-1)^5] \). Correct? Start by making \( \frac{d(log(u))}{dx}\frac{ du}{dx}\), for \( u = e^{16} (3x-1)^4 (x^3-1)^5 \). That's apply the chain rule.

Answer 2

Ahh I see

Answer 3

Afterwards, the constants (e^16) should cancel out.

Answer 4

And then you will have to use the product rule :-)

Answer 5

And, there is a typo there, it should be \(\Large \frac{d(\ln{u})}{du} \)

Answer 6

Did you give it a shot? :-)

Answer 7

yea

Answer 8

i got 4ln(3x-1)+5ln(x^3-1)

Answer 9

That does not look correct. The ln should disappear. Remember that d(log(u))/du = 1/u.

Answer 10

I got \[ \frac{12(3x -1)^3(x^3 - 1)^5 + 5 (3x-1)^4(3x^2)(x^3 - 1)^4}{(3x−1)^4(x^3−1)^5}\]

Answer 11

r u sure of this?..my teacher just added on y=4x^2 to the equation i just posted..

Answer 12

http://www.wolframalpha.com/input/?i=d%28ln%5B%28e+%5E%284%5E2%29%29+%283x-1%29%5E4+%28x%5E3-1%29%5E5%5D%29%2Fdx Hmm.

Answer 13

If you click show steps, it does coincide with my answer. I don't see how we can get ln out of this.

Answer 14

the original equation should be
d/dx of ln[(e ^(4x^2)) (3x-1)^4 (x^3-1)^5]

Answer 15

\[\frac{d( \ln[(e ^{4x^2} (3x-1)^4 (x^3-1)^5] )}{dx}\]?

Answer 16

yea.. but the question was to find the derivative, i just used d/dx to signify that

Answer 17

Yup. Still, I don't see a ln here... @TuringTest can you help me, mate? :-)

Answer 18

?

Answer 19

I still think that it should not have a log. http://www.wolframalpha.com/input/?i=d%2Fdx+of++ln%5B%28e+%5E%284x%5E2%29%29+%283x-1%29%5E4+%28x%5E3-1%29%5E5%5D

Answer 20

Andesa bmp is right. the derivative of ln(x) is 1/x just imagine everything in the brackets after the ln as x and it will be clear why ln shouldn't be in your final answer.

(1/(e^16(3x−1)^4(x^3−1)^5))* the derivative of what's in the brackets
understand?

Answer 21

Of course bmp right! ( lnu )' = u'/ u
That's how you take derivative of ln

Answer 22

Let u = e ^(4^2)) (3x-1)^4
-> u' =
Let v = (x^3-1)^5
-> v' =
y' = U'/ U = ( u'v + uv' ) / U

Answer 23

okay folks thanks...majority wins...