Question

ap calc ab help
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Answer 1

@misty1212

Answer 2

\[\frac{d}{dx}(g(f(x))) = g'(f(x)) \cdot f'(x)\]

Answer 3

HI!!

Answer 4

and hi @Jhannybean !!

Answer 5

Hi!!! @misty1212

Answer 6

Hi everyone xD Thanks for helping me!!

Answer 7

you have all the numbers you need to plug in to what @Jhannybean wrote above
plug them in, see what you get

Answer 8

Wait so I understand how to plug in the f(x) and f'(x) but how do I do the g(x) and g'(x)

Answer 9

let me just add one little bit \[\frac{d}{dx}(g(f(3x))) = g'(f(3x)) \cdot f'(3x)\cdot 3\]

Answer 10

So now we have \(g'(f(x)) \cdot f'(x)\) take it one step at a time. \[g'(f(1)) \cdot f'(1)=~?\]

Answer 11

careful a little
it is \(f(3x)\) so at \(x=1\) it is \(f(3)\)

Answer 12

Oooo stupid me. I misread a portion of the problem.

Answer 13

Yeah. Just noticed.

Answer 14

So if x=1 at f(3x) = f(3) what am i supposed to use to plug in.. im actually really confused

Answer 15

now got to read them off of the table
at \(x=1\) it is \[g'(f(3)) \cdot f'(3)\cdot 3\]

Answer 16

what is \(f'(3)\) from the table?

Answer 17

2

Answer 18

i mean 10

Answer 19

and what is \(f(3)\) from the table?

Answer 20

2

Answer 21

and finally, what is \(g'(2)\)?

Answer 22

5

Answer 23

So would it be 5(10)(3)=150?

Answer 24

Yeah

Answer 25

number therapy for learning the chain rule

Answer 26

Haha thank you so much! Okay so would that be how you do it at x=1 or is that just an example from the chart?

Answer 27

We did it at x=1

Answer 28

Oh okay i understand so we just broke it up into components

Answer 29

Thank you so much both of you for your help!!

Answer 30

\[g'(f(1))\cdot f'(1) = \\ g'(f(3(1))) \cdot f'(3(1)) \cdot 3 = \\ g'(f(3)) \cdot f'(3) \cdot 3 = \\ g'(2) \cdot 10\cdot 3 = \\ 5 \cdot 10\cdot 3 = 150\]