Question

take the derivative of 1/7-x

Answer 1

\[\frac{d}{dx}\frac{1}{7-x}\]?

Answer 2

first one

Answer 3

= (1/7)x^-1 -(1/7)x^-2 or 1/(7+x^2)

Answer 4

(1/7)x^-2 disregard the middle

Answer 5

why is there negative 2?

Answer 6

x^-2 is the same as 1/x^2

Answer 7

let me try to do it right.
the derivative of
\[\frac{1}{f}\] is
\[-\frac{f'}{f^2}\] so here you have
\[f(x)=7-x\]
\[f'(x)=-1\] and your answer is
\[\frac{1}{(7-x)^2}\]

Answer 8

where did the square on the denominator come form? isn't when you take the derivative of something you have to subtract the exponent?

Answer 9

the square came from the fact that the derivative of
\[\frac{1}{f(x)}=-\frac{f'(x)}{f^2(x)}\]

Answer 10

what happen to the negative for second equation?

Answer 11

if you want to use exponential notation that will work too. it is more cumbersome though. you write
\[\frac{1}{7-x}=(7-x)^{-1}\] so
\[\frac{d}{dx}\frac{1}{7-x}=\frac{d}{dx}(7-x)^{-1}=-1(7-x)^{-2}\times -1=(7-x)^{-2}=\frac{1}{(7-x)^2}\]

Answer 12

ahh the -1 is from the "chain rule"

Answer 13

the derivative of
\[7-x\] is -1

Answer 14

oh ok, thanks could you check my graph problem that you couldn't see?

Answer 15

i remember it. you had a parabola that faced down

Answer 16

yes

Answer 17

I attached another document, hopefully its clearer

Answer 18

the function is increasing and then decreasing, so your derivative will be positive (above the x - axis) then negative (below)

Answer 19

it will be 0 where your function changes direction, at the maximum in this case

Answer 20

looks like #1 from the picture

Answer 21

yeah, its also can of look like number 5

Answer 22

yes it could but your function if i am reading the graph correctly changes direction at 4
so your derivative should be 0 at 4