Question

Please help

Suppose that f(pi/3)=4 and f'(pi/3)=-2 let g(x)=f(x)sinx and h(x)=cosx/f(x). Find g'(pi/3) and h'(pi/3)

Answer 1

Have you applied product rule to differentiate g(x)=f(x)sin(x)?

Answer 2

the other one you could use quotient rule

Answer 3

Not yet. Do I put (pi /3) in for x?

Answer 4

after differentiating

Answer 5

I'm a bit confused on how to differentiate it

Answer 6

do you know the product rule?

Answer 7

Yes

Answer 8

so can you try to apply it here:
(f(x)*sin(x))'=?

Answer 9

you do know the product rule is:
(u*v)'=u'*v+v'*u

Answer 10

?

Answer 11

replace u with f(x)
and replace v with sin(x)

Answer 12

F'(x)•sinx+f(x)•cosx?

Answer 13

\[g(x)=f(x)\sin(x) \\ g'(x)=f'(x)\sin(x)+f(x)\cos(x) \\ \text{ replace } x \text{ with } \frac{\pi}{3} \\ \text{ to find } g'(\frac{\pi}{3})\]

Answer 14

Would the derivative of (pi/3) be 0?

Answer 15

yes but why are you finding the derivative of pi/3

Answer 16

When you replace it for x in the problem don't you have to find the derivatives?

Answer 17

you already differentiated the problem ...

Answer 18

you are finding g'(pi/3) by replacing x with pi/3

Answer 19

look at your problem it says to find g'(pi/3) when f(pi/3)=4 and f'(pi/3)=-2

Answer 20

\[g(x)=f(x)\sin(x) \\ g'(x)=f'(x)\sin(x)+f(x)\cos(x) \\ \text{ replace } x \text{ with } \frac{\pi}{3} \\ \text{ to find } g'(\frac{\pi}{3})\]

\[g'(\frac{\pi}{3})=f'(\frac{\pi}{3}) \sin(\frac{\pi}{3})+f(\frac{\pi}{3}) \cos(\frac{\pi}{3})\]
can you continue
it is just pluggin numbers given...and following order of operatios

Answer 21

I ended up with 2sin(pi /3)+cos(pi/3)

Answer 22

why don't you replace f'(pi/3) with -2 since f'(pi/3) is -2
and replace f(pi/3) with 4 since f(pi/3)=4

also cos(pi/3) and sin(pi/3) should be known by you
cos(pi/3)=1/2 and sin(pi/3)=sqrt(3)/2

Answer 23

So g'(pi /3)= sqrt 3/2 ?

Answer 24

\[g'(\frac{\pi}{3})=f'(\frac{\pi}{3}) \sin(\frac{\pi}{3})+f(\frac{\pi}{3}) \cos(\frac{\pi}{3}) \\ g'(\frac{\pi}{3})=-2 \cdot \frac{\sqrt{3}}{2}+4 \cdot \frac{1}{2} \\ g'(\frac{\pi}{3})=-\sqrt{3}+2 \text{ this is not the same as } \frac{\sqrt{3}}{2}\]

Answer 25

all I could really do is do -2/2 and 4/2
which is -1 and 2 respectively

Answer 26

Oh okay I get it

Answer 27

are you sure?
I'm worried you have forgotten order of operations...or adding like terms (and not adding non-like terms)

Answer 28

I guess we will see in the next problem
I suggest using quotient rule for the next one you have there

Answer 29

Okay so since h(x)=cosx/f(x) that means g(x)= cosx ?

Answer 30

g(x) was the previous function we worked with
g(x)=f(x)sin(x) we already found g'(pi/3)

Answer 31

to differentiate h(x)=cos(x)/f(x) you do not need the previous function
just the quotient rule

Answer 32

I meant as y=f(x)/g(x)

Answer 33

\[(\frac{\text{top}}{\text{bottom}})'=\frac{(\text{top})' \cdot \text{ bottom }- (\text{bottom})' \cdot \text{ top}}{(\text{bottom})^2}\]

Answer 34

Okay so I got sin(pi/3)•f(pi/3)-cos(pi/3) / pi/3^2

Answer 35

derivative of cos(x) is -sin(x)
and seems you forgot to write f'(x) .
Also why is the bottom pi/3^2
and not (f(pi/3))^2

Answer 36

\[(\frac{\cos(x)}{f(x)})'=\frac{(\cos(x))' \cdot f(x)-(f(x))' \cdot \cos(x)}{(f(x))^2} \text{ all I did was plug into } \\ \text{ quotient rule about }\]

Answer 37

\[(\cos(x))'=-\sin(x) \\ (f(x))'=f'(x)\]

Answer 38

So I have now -sqrt3/2•4-(-2)(1/2) over 16

Answer 39

this right:
\[\frac{-\frac{\sqrt{3}}{2} \cdot 4 - (-2) \cdot \frac{1}{2}}{16} ?\]

Answer 40

you can simplify this a little

Answer 41

-3 sqrt 3 /16

Answer 42

-1 sqrt 3/16 ***

Answer 43

how did you get that...

Answer 44

on top you have multiplication(division) and subtract
you are suppose to do the multiplication first
\[- \frac{\sqrt{3}}{2} \cdot 4 = - \frac{4}{2} \sqrt{3} =? \\ -(-2) \cdot \frac{1}{2}=?\]

Answer 45

Wouldn't it be -2sqrt3 +1?

Answer 46

yes that is correct for the top

Answer 47

and there is nothing more you can do to simplify the top

Answer 48

\[\frac{-\frac{\sqrt{3}}{2} \cdot 4 - (-2) \cdot \frac{1}{2}}{16} \\ \frac{\frac{ -\sqrt{3}}{2} \cdot 4 +-(-2) \cdot \frac{1}{2}}{16} \\ \frac{ \color{red}{-\frac{\sqrt{3}}{2} \cdot 4} + \color{blue }{-(-2) \cdot \frac{1}{2}}}{16} = \frac{ \color{red}{-2 \sqrt{3}}+\color{blue}{1}}{16} \]

Answer 49

if you had
\[-2 \sqrt{3}+1\sqrt{3} \text{ this would be } -1 \sqrt{3} \text{ or } -\sqrt{3}\]
but you do not have like terms you can go no further

Answer 50

Okay thanks for the help!

Answer 51

np