Question

serious help with a calc problem!medal!

Answer 1

use chain rule :
\(h(x) =g[f(3x)] \\ h'(x) = ... ?\)

Answer 2

example :
\(\Large j(x) = f[(k(4x+9))] \\ \Large j'(x) = 4f'[k(4x+9)]k'(4x+9)\)

Answer 3

3f'(3x)g'(f(3x))

Answer 4

yesh
now to get
\(\Large h'(1)\)

plug in x=1

Answer 5

in your answer

Answer 6

h?

Answer 7

whats your doubt?

Answer 8

ohh! didnt see the second thing you wrote

Answer 9

to find \(h'(1) \\ from ~~h'(x) \\ plug ~ in~ x=1\)

Answer 10

okk :)

Answer 11

3f'(3)g'(f(3))

Answer 12

now get f(3) from the table

Answer 13

f(3) is in f(x) when x=3

Answer 14

i plug in 3 for f?

Answer 15

you see f(x) row

get the value in the column where x=3

that gives you f(3)

Answer 16

I think, that:
\[h'(1)=\lim _{f(3x)\rightarrow f(3)}\frac{ g(f(3x))-g(f(3)) }{ f(3x)-f(3) }*\]
\[*\lim _{y \rightarrow 3}\frac{ f(y)-f(3) }{ y-3 }*\lim _{x \rightarrow1}\frac{ 3x-3 }{ x-1 }=...\]

Answer 17

which is 2 right? @hartnn

Answer 18

yes
f(3) = 2

Answer 19

still wondering what michele did

Answer 20

from the same table find
f'(3) too

Answer 21

1

Answer 22

I have applied the definition of derivative of composite function, and that I change variable in limit operations

Answer 23

3f'(3)g'(f(3))

= 3 (1) g'(2) = ..

whats g'(2) =

Answer 24

5(:

Answer 25

= 3 (1) g'(2) = 3(1)(5) = 15

Michele, you get 15 too?

Answer 26

which is equal to my result, since:
\[h'(1)=g'(f(3))*f'(3)*3=5*1*3=15\]

Answer 27

same answer , different method(: thank you guys!!

Answer 28

thanks for the alternate method Michele :D

Answer 29

thanks! @hartnn @mondona

Answer 30

:)