Question

find f'(x). f(x)=ln(sin(pi*x^2)).

Answer 1

Do you remember the chain rule? :-)

Answer 2

yes, think so.

Answer 3

apply it... :D

Answer 4

Then, start from the outside. What is the derivative of ln(u)?

Answer 5

(1/u) * u' ?

Answer 6

Correct. But now notice that u' here is sin(x^2 pi). Do the chain rule again, what is the derivative of sin(u)?

Answer 7

Or, I should be more careful, the derivative of sin(v). We don't want any confusions with the prior u.

Answer 8

cos(v) * v'

Answer 9

Now, what is v'?

Answer 10

2pix ?

Answer 11

Yup. Well done. Now plug in back for u and v. u is sin(x^2pi) and v is x^2 pi, and you are done.

Answer 12

what was the first u?

Answer 13

the sin one u just said?

Answer 14

Yup, remember, ln(u), so everything inside of the ln is the first u.

Answer 15

thank you!

Answer 16

No problem :-)

Answer 17

did you get the correct answer?

Answer 18

yes, ty

Answer 19

ok

Answer 20

good luck now

Answer 21

( 1/sin(pi*x^2) ) * cos(pi*x^2) * 2pi*x

Answer 22

That's correct. Or you can rewrite it as \( \cot{\pi x^2} (2\pi x) \), since cos/sin is cot. But your answer is correct :-)

Answer 23

is that considered more simplified?

Answer 24

I guess so; if you are asked to enter in simplified form in an online answer, I would say yes.

Answer 25

nah just says find f'(x)

Answer 26

then i'm suppose to f'(1/2) but i think i just plug it in for x

Answer 27

Yup, do that :-)

Answer 28

does ti-84plus have cot?

Answer 29

guess i have to do 1/cofunction

Answer 30

Sorry mate. You can use www.wolframalpha.com to do the calculation.

Answer 31

kk

Answer 32

What did you get? :-)

Answer 33

pi

Answer 34

Correct :-)

Answer 35

woot