Question

hhhheeeellpppp! Calculus! Evaluate the derivative at of the function at the indicated point. f(x)=1/(x^2-3x)^2 (4, 1/16)

Answer 1

use chain rule
let u=x^2-3x
du/dx= .....

Answer 2

okay thanks!

Answer 3

next, y=u^2
dy/du= .......
multiply of both to get dy/dx

Answer 4

hello! you got this?

Answer 5

not REALLY

Answer 6

HELOO!

Answer 7

ok as usual we go slow
if you had\[\frac{1}{x^2}\] you would know what to do, right?

Answer 8

yes?

Answer 9

you mean "yes!" or "yes?"

Answer 10

yes?

Answer 11

ok so i will take that as a maybe
if you just had \(\frac{1}{x^2}\) you could use the power rule, after writing \[\frac{1}{x^2}=x^{-2}\]

Answer 12

yes!!!!

Answer 13

then by the power rule, the derivative would be ...

Answer 14

-2x^-3

Answer 15

right, also known as \(-\frac{2}{x^3}\)

Answer 16

yes :}

Answer 17

BUT you do not have \(\frac{1}{x^2}\) you have
\[\frac{1}{(x^2-3x)^2}\] so you have to adjust

Answer 18

kay:l

Answer 19

so here is what you do
take your previous answer, \(-\frac{2}{x^3}\) and replace \(x\) by \(x^2-3x\) so instead of getting
\[-\frac{2}{x^3}\] you get
\[\frac{-2}{(x^2-3x)^3}\]

Answer 20

THEN multiply this by the derivative of \(x^2-3x\)

Answer 21

let me know if you have a question

Answer 22

u confused the s%&# outta me boy or girl satellite 73

Answer 23

ho ho ho i see i am good at that

Answer 24

thanks so much encouragement :p

Answer 25

what is the derivative of \(\sin(x)\) ?

Answer 26

cosx

Answer 27

ok good
now what is the derivative of \(\sin(x^2)\) ?

Answer 28

man ummmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm cosx^2????

Answer 29

hmmmmmmmmmmmmmmmmm
nope

Answer 30

you have to take \(\cos(x^2)\) and multiply it by the derivative of \(x^2\) which is \(2x\)

Answer 31

this is called the "chain rule" because you have a chain of relations or functions. we say for \(\sin(x^2)\) the "inside" function is \(x^2\) and the "outside" function is \(\sin\)

Answer 32

so you have a composition of functions

Answer 33

so it'll be 2xcosx?

Answer 34

no

Answer 35

YEAH!!!!! yaee!

Answer 36

wait what?

Answer 37

don't change the inside piece, it is still \(x^2\)

Answer 38

o okay 2xcosx^2?

Answer 39

the answer would be \(2x\cos(x^2)\)

Answer 40

right
in english: the derivative of a composite function is:
the derivative of the "outside" function, evaluated at the inside function,
times the derivative of the "inside" function

Answer 41

so for example, since the derivative of sine is cosine, then the derivative of \(\sin(x^2)\) is
\[2x\cos(x^2)\]

Answer 42

how about the derivative of \((3x^2-x)^{10}\) ?

Answer 43

here the "inside" function is \(3x^2-x\) while the "outside" function is "to the tenth power""

Answer 44

its aaaaaaaaaaaaaaa........................ hmmmmmmmmmummmmmmmmmmmmmmmmmmm.............. 10(3x^2-x)
10(6x-1)=======60x-1! :}

Answer 45

NO! 60x-10

Answer 46

:]

Answer 47

close, very close
it is \[10(3x^2-x)^9\times (6x-1)\]

Answer 48

you forgot that the derivative of something to the power of ten is ten something to the power of nine by the power rule

Answer 49

cuz its u*u(prime) right?

Answer 50

mmk

Answer 51

you forgot the 9th power

Answer 52

lets try this
\[(5x^2-3)^6\]

Answer 53

last one cuz i gotta go satellite 73 (sorry :[ )
6(5x^2-3)^5=========6(10x-3)^5======(60x-18)^5?????

Answer 54

it is
\[6(5x^2-3)^5\times (10x)\]

Answer 55

you cannot distribute because of that exponent

Answer 56

but that is the idea

Answer 57

so you have
\[(x^2-3x)^{-2}\]
derivative is
\[-2(x^2-3x)^{-3}\times (2x-3)\]

Answer 58

ok :]

Answer 59

bye~

Answer 60

good luck
bye