Question

Find the equation of the tangent line to f(x)=1/x^-2 at x=9

Answer 1

can you clarify the equation?

Answer 2

i mean 1 over square root of x

Answer 3

You need to find the derivative of 1/x first.

Answer 4

i know and i forgot how to do it the long way

Answer 5

oh it is 1/sqrt(x)

Answer 6

you have to do it the long way ?

Answer 7

lol i forgot also....

Answer 8

Yes. WE havent learned the shortcut yet but i have

Answer 9

\[f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]

Answer 10

i tried that and i got stuck after plugging it in

Answer 11

\[f'(x)=\lim_{h \rightarrow 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h} =\lim_{h \rightarrow 0}\frac{1}{h} \cdot (\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}})\]

Answer 12

First step : Combine fractions in the ( )

Answer 13

Second step: Combine like terms on top

Answer 14

Third step: Cancel a factor of h on bottom with the factor of h on top.

Answer 15

Last step: Replace remaining h's with 0.

Answer 16

the combined fractions for the first step are sqrt x - sqrt x+h all over sqrtx * sqrt x+h

right?

Answer 17

oh i missed step rationalize numerator

Answer 18

That should be first and half step lol

Answer 19

how would i do that

Answer 20

\[f'(x)=\lim_{h \rightarrow 0}\frac{1}{h}(\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x+h} \sqrt{x}}) \cdot \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}\]
multiply by top's conjugate on top and bottom

Answer 21

do not distribute on bottom
just on top

Answer 22

what would remain after that?

Answer 23

multiply the top and see

Answer 24

recall (a-b)(a+b)=a^2-b^2

Answer 25

x-x+h?

Answer 26

close
x-(x+h)

Answer 27

oh ok yeah I figured

Answer 28

and then thats over the current denominator
of sqrtx+h * sqrtx

Answer 29

how would i fix up the bottom?

Answer 30

h/h=1

Answer 31

what? how does h/h fix the sqrts
on the bottom

Answer 32

why do the sqrt need to be fixed?

Answer 33

so its (1/h) * (x-x+h)/ (sqrt x+h)(sqrtx)

Answer 34

again we have x-(x+h) on top you are also missing the other part of the bottom

Answer 35

x-x-h=-h

and again h/h=1

Answer 36

the bottom part of the congragate?

Answer 37

\[f'(x)=\lim_{h \rightarrow 0}\frac{1}{h}(\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x+h} \sqrt{x}}) \cdot \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} \\ =\lim_{h \rightarrow 0}\frac{1}{h}\frac{x-(x+h)}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}\]

Answer 38

yeah

Answer 39

x-x-h=-h
h/h=1

Answer 40

You are on that third step I mentioned

Answer 41

so the bottom cancels out?

Answer 42

:(

Answer 43

doesn't x-(x+h)=-h?

Answer 44

yes

Answer 45

do you see a factor h on bottom?

Answer 46

oh dear god

Answer 47

idk what i was thinking

Answer 48

ok yeah now i completely get it i guess i was trying to solve the problem and get a number rather than just find the derivative

Answer 49

\[\lim_{h \rightarrow 0}\frac{- \cancel{h}}{\cancel{h} \sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}\]

Answer 50

now you can replace the h's with 0

Answer 51

Ok well thank you and i apologize for my being dumb for about 20min there lol

Answer 52

It's okay. I was eating oreos part ways through it.

Answer 53

is there a way i "friend" you for help i might need in the future?

Answer 54

friend you is a cool term
There is fanning
You can message me and see if i'm on when you need help

Answer 55

I will respond if I'm on

Answer 56

Thank you