Question

Suppose that
x(f(x))^3 + xf(x) = 30,
and that f(3)=2. Find f'(3).
f'(3) =

Answer 1

can you differentiate the given equation?

Answer 2

the x(f(x))^3+xf(x)=30

Answer 3

\[\frac{d}{dx}(f(x))^3 \text{ can be found by chain rule }\]

Answer 4

do you know how to do that?

Answer 5

\[3f(x)^2*f'(x)\]

Answer 6

that's good

Answer 7

\[\frac{d}{dx} x(f(x))^3 = (f(x))^3 \frac{d}{dx} x+ x \frac{d}{dx}(f(x))^3\]

Answer 8

and you already found the derivative of (f(x))^3

Answer 9

\[\frac{d}{dx} x(f(x))^3 = (f(x))^3 \frac{d}{dx} x+ x \frac{d}{dx}(f(x))^3 \\ =(f(x))^3 \frac{dx}{dx}+x 3(f(x))^2f'(x)\]

Answer 10

So you do know I just used the product rule right?

Answer 11

You can apply the product rule to also differentiate xf(x)

Answer 12

Anyways once you have differentiate both sides
replace all the x's with 3 since you attempting to find f'(3) and are given f(3)

Answer 13

Okay i got a little lost can i just put in the 3s in here? \[f(x)^3+3xf(x)^2*f'(x)+f(x)+xf'(x)=0\]

Answer 14

yes replace all the little x's you see with 3

Answer 15

\[f(3)^3+3(3)f(3)^2*f'(3)+f(3)+3f'(3)=0\]

Answer 16

and you f(3) is 2

Answer 17

\[2^3+3(3)2^2f'(3)+2+3f'(3)=0\]

Answer 18

solve that equation for f'(3)

Answer 19

you may want to simplify first

Answer 20

ok i got -11.5 but I think that's wrong.

Answer 21

I think that might be wrong too

Answer 22

I have answer between -1 and 0

Answer 23

and -11.5 doesn't fit in there

Answer 24

but I could have done something wrong too

Answer 25

\[2^3+3(3)2^2f'(3)+2+3f'(3)=0 \\ 8+36f'(3)+2+f'(3)=0\]

Answer 26

combine like terms

Answer 27

oops left off the 3 in front of the f'(3)

Answer 28

\[2^3+3(3)2^2f'(3)+2+3f'(3)=0 \\ 8+36f'(3)+2+3f'(3)=0\]

Answer 29

ok combine like terms

Answer 30

8+2=?
36f'(3)+3f'(3)=?

Answer 31

oops I see what I did wrong. -10/39 is the answer.

Answer 32

\[10+39f'(3)=0\]

Answer 33

yah!

Answer 34

thanks so much

Answer 35

np