Question

derivatives

Answer 1

Find the derivative of f(x) = 5/x at x = -1.

Answer 2

hint : \(\dfrac{5}{x} = 5x^{-1} \)

Answer 3

hint2 : \(\dfrac{d}{dx} (x^n) = nx^{n-1}\)

Answer 4

just use the above formula, unless you want to do it by definition (first principles)...

Answer 5

wait... so i put -1 in for x? @ganeshie8

Answer 6

first take the derivative

Answer 7

\(f(x) = 5/x = 5x^{-1}\)


\[\dfrac{d}{dx}(f(x)) = \dfrac{d}{dx}(5x^{-1} ) = 5 \dfrac{d}{dx}(x^{-1} ) = 5(-1*x^{-1-1}) = -5x^{-2} = -\dfrac{5}{x^2}\]

Answer 8

plugin x = -1

Answer 9

holy crap what is that o.o

Answer 10

\[\dfrac{d}{dx}(f(x)) \Bigg|_{x=-1 }= -\dfrac{5}{x^2} \Bigg|_{x=-1 } = ?\]

Answer 11

oh my gosh... what is this /).(\

Answer 12

i think you want to do it from first principles since you're bit surprised by the d/dx notation

Answer 13

\[f'(x) = \lim \limits_{x\to a} \dfrac{f(x) - f(a) }{x-a}\]

Answer 14

seen this before ?

Answer 15

* corrected : \[f'(a) = \lim \limits_{x\to a} \dfrac{f(x) - f(a) }{x-a}\]

Answer 16

yeah c:

Answer 17

cool :) evaluate `f(a)` and plug the value in above formula

Answer 18

it would be -1 right? c:

Answer 19

yes \(a\) would be \(-1\) since you want the derivative at \(-1\)

Answer 20

\(f(x) = \dfrac{5}{x}\)

\(f(a) = f(-1) = ?\)

Answer 21

wouldnt that just equal the same thing i started with? itd still be 5/x-1 :c

Answer 22

\(f(x) = \dfrac{5}{x}\)

\(f(a) = f(-1) = \dfrac{5}{-1} = -5\)

right ?

Answer 23

plug that value in ur derivative formula :

\[f'(a) = \lim \limits_{x\to a} \dfrac{f(x) - f(a) }{x-a}\]

\[f'(-1) = \lim \limits_{x\to -1} \dfrac{\frac{5}{x} - (-5) }{x-(-1)}\]

Answer 24

fine so far ?

Answer 25

its almost done, u just need to simplify, cancel stuff and take the limit

Answer 26

but you said x=5/x

Answer 27

??

Answer 28

idk.... well itd be 5/x +5 /x+1

Answer 29

you're right !

Answer 30

wait then what do i do?

Answer 31

\[f'(-1) = \lim \limits_{x\to -1} \dfrac{\frac{5}{x} - (-5) }{x-(-1)} = \lim \limits_{x\to -1} \dfrac{\frac{5}{x} +5 }{x+1} = \lim \limits_{x\to -1} \dfrac{5(1+x) }{x(x+1)} \\ = \lim \limits_{x\to -1} \dfrac{5}{x} = \dfrac{5}{-1} = -5 \]

Answer 32

complete work above^
let me knw if smthng doesnt make sense..

Answer 33

okay so id just that the 5 and multiply it in order to cancel it out c: makes sense thank you c:

Answer 34

yup :) i have pulled the 5 out and and multiplied x both top and bottom,
added more steps in between :


\[f'(-1) = \lim \limits_{x\to -1} \dfrac{\frac{5}{x} - (-5) }{x-(-1)} = \lim \limits_{x\to -1} \dfrac{\frac{5}{x} +5 }{x+1} \\ = \color{red}{ \lim \limits_{x\to -1} \dfrac{5\left(\dfrac{1}{x}+1\right) }{x+1} = \lim \limits_{x\to -1} \dfrac{5\left(\dfrac{1+x}{x}\right) }{x+1} } = \lim \limits_{x\to -1} \dfrac{5(1+x) }{x(x+1)} \\ = \lim \limits_{x\to -1} \dfrac{5}{x} = \dfrac{5}{-1} = -5

\]

Answer 35

do you think you could help me with some more of these if i post separate questions

Answer 36

il try :) but u better do most of the work as you're familiar wid the method now :P

Answer 37

all we need to do is plug it into the formula, rearrange it somehow to cancel the bad denominator