Question

Write the first derivative of the given function.

f(x) = x^2[ln(x)]^18

I got: f'(x) = 2x(lnx)^18 + 18x(lnx)^17

Now part 2 asks, For which x-values does f(x)
have horizontal tangents? I'm not getting it right so I need your help :) please and thank you!

Answer 1

I think your first derivative is wrong. Try to do the product rule again. Remember that you will have to do the chain rule for (lnx)^18

Answer 2

Oh god, you got the problem that I struggled for 1 week to solve... (they had us get the f13)

anyways.

x^2[ln(x)]^18
a = x^2
b = [ln(x)]^18
a' = 2x
b' = [ln(x)]^17*(18/x)

f' = 2x([ln(x)]^18) + (x^2)([ln(x)]^17*(18/x))

horizontal = 0 = f'
0=2x([ln(x)]^18) + (x^2)([ln(x)]^17*(18/x))
x = 1

Answer 3

it's not 1. what does your derivative looks different from mine?

Answer 4

ah wait, entered it wrong...
2 x log^17(x) (9+log(x))

x1 = 1
x2 = 1/e^9

Answer 5

i realize that to find horizontal tangent you put f'(x) = 0. but it's so complicated. do you find the limit as it approaches 0. is there a method?

Answer 6

@kimmy the second part of your answer is incorrect

its 18(lnx)^17 * 1/x * x^2 = 18x (lnx)^17

- excuse me - its correct!!

Answer 7

for horizontal tangents f'(x) = 0

Answer 8

dreadslicer got the answer right? but how??

Answer 9

Once you have the correct derivative, you need to set it equal to zero. Every solution will be a horizontal tangent.

Answer 10

but using the derivative i provided (which is correct), how could you find a solution with ln? i only got x=1

Answer 11

x can't be 0 because ln has to be greater than 0. that's why i'm not sure how he got 1/e^9