Question

if f(x)=1/x-3 and g(x)=f'(x) find g'(x)

Answer 1

this is asking for the inverse of the inverse. This is probably f(x) logically.

Start by finding the inverse of f(x)

Answer 2

so g'(x) = f'' (x)

Answer 3

I might actually be wrong. Anyhow.
\[y = \frac{1}{x-3}\]
\[x = \frac{1}{y-3}\]
solve for y

Answer 4

inverse or derivative?

Answer 5

derivative. it is the derivative of the derivative. I got it. thx. I have another question exactly the same except g(x)= f^-1 (x)

Answer 6

it's asking for 2rd derivative of f

Answer 7

she already got it :)

Answer 8

so what do we need to find?
g'(x)?

Answer 9

or g(x)?

Answer 10

if f(x)=1/x-3 and g(x)=f^-1 (x) find g'(x)

Answer 11

So lets find the inverse and follow what bibby wrote above

Answer 12

\[y = \frac{1}{x-3}\]
Then we switch our x and y
\[x = \frac{1}{y-3}\]
Then Solve for y
\[y=\frac{1}{x}+3\]
so \(g(x)=\frac{1}{x}+3\)

Answer 13

Now find the derivative of g(x)

Answer 14

\[g'(x)=-\frac{1}{x^2}\]

Answer 15

remember that you can add derivatives so \[g'(x) = \frac{d}{dx}\frac{1}{x} +\frac{d}{dx}3\]

Answer 16

1/3x^2???

Answer 17

the derivative of 3 (a constant) is 0

Answer 18

you can use either the power rule because 1/x = x^-1 or the division rule

Answer 19

1/x^2?

Answer 20

-1/(x^2)

\[x^n - > nx^{n-1}\]
\[\frac{ d }{ dx }x^{-1}\]

Answer 21

okay

Answer 22

ok like what dont u follow?