Question

f(x) is a continuous and differentiable function. The table below gives the value of f(x) and f'(x) at several values.

Answer 1

\[\Large g(x)\quad=\quad \frac{1}{f(x)}\]Hmm this one is a little tricky.
Have you at this point learned some of the basic derivative rules like `power rule`?

Answer 2

Yes, I have

Answer 3

Ok then using rules of exponents, let's write our relationship like this:\[\Large g(x)\quad=\quad \left[f(x)\right]^{-1}\]

Answer 4

From here, we can apply the power rule and then chain rule.

Answer 5

No pluggin in?

Answer 6

No, we'll want to take a derivative before plugging :)

Answer 7

\[g(x)=-f(x)^{-2}\]

Answer 8

Good, but don't forget the chain rule!\[g(x)=-f(x)^{-2}\color{royalblue}{f'(x)}\]

Answer 9

Woops we got a little sloppy there and forgot our prime on the g,\[\Large g'(x)=-f(x)^{-2}\color{royalblue}{f'(x)}\]

Answer 10

Mmm, is that all for the chain rule?

Answer 11

Should I do something else?

Answer 12

No that should be fine :)
We took the derivative of the inner function f(x).
If the function has been something like f(2x), yes we would .. blah let's not think about that right now. It'll just confuse you maybe.

Answer 13

Ok, so now do we plug in?

Answer 14

Yes good! :)\[\Large g'(\color{orangered}{2})=-f(\color{orangered}{2})^{-2}f'(\color{orangered}{2})\]

Answer 15

And our table should come in very handy at this point!

Answer 16

\[g'(2)=-(-6)^{-2}(-2)\]
\[g'(2)=-\frac{ 1 }{ 36 }(-2)\]
\[g'(2)=\frac{ 1 }{ 18 }\]

Answer 17

Am I right?

Answer 18

@zepdrix

Answer 19

Yayyy good job \c:/

Answer 20

Alright, thank you again @zepdrix !

Answer 21

np!