Question

f is a differentiable function on the interval [0, 1] and g(x) = f(4x). The table below gives values of f ′(x). What is the value of g ′(0.1)?

Answer 1

https://gyazo.com/b435133c157b25d1c60f85c59cdeba34

Answer 2

Just multiplied by 4. Would that be right?

Answer 3

@zepdrix it's destiny telling you to help me ;P

Answer 4

hehe

Answer 5

\[\large\rm g(x)=f(4x)\]Take derivative, apply chain rule on the right side of the equation.

Answer 6

Understand that step? :d

Answer 7

Chain rule on th eright side?

Answer 8

\[\large\rm \frac{d}{dx}f(4x)\quad=f'(4x)\cdot(4x)'\]Yes, chain rule.
Multiply by the derivative of the inner function.

Answer 9

\[\large\rm g(x)=f(4x)\]Differentiate with respect to x,\[\large\rm \frac{d}{dx}g(x)=\frac{d}{dx}f(4x)\]Leading to\[\large\rm g'(x)=f'(4x)\cdot(4x)'\]this\[\large\rm g'(x)=f'(4x)\cdot(4)\]

Answer 10

Confused?

Answer 11

I got it just confused on how that'll help me.

Answer 12

Well we're looking for g'(0.1).
Let's plug that into this relationship.\[\large\rm g'(0.1)=4\cdot f'(4\cdot0.1)\]\[\large\rm g'(0.1)=4\cdot f'(0.4)\]

Answer 13

And then use the table to finish the problem :o

Answer 14

Ohhhhhhhhhhhhhh
Got it
-16

Answer 15

Yay!

Answer 16

8/8 m8 that was gr8 thx

Answer 17

lol

Answer 18

Good job m8