Question

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

Answer 1

by hand i bet

Answer 2

I worked it out to [(5/x+h)-(5/x)]/h but I'm not sure if the 5 and x cancel out or not.

Answer 3

\[f(x)=\frac5x\]
Then :
\[f'(x)=-\frac{5}{x^2}\]
So :
\[f'(-1)=-\frac{5}{(-1)^2}=-5\]

Answer 4

Shouldn't you use the derivative equation

Answer 5

@Noura11

Answer 6

\[\lim_{x\to -1}\frac{\frac{5}{x}+5}{x+1}
\]

Answer 7

I don't know how to do that. What did you do there?

Answer 8

if you are doing this by hand, not by shortcut methods, then
\[\f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]

Answer 9

since \(f(-1)=-5\) in your case you get
\[f'(-1)=\lim_{x\to -1}\frac{\frac{5}{x}+5}{x+1}\]

Answer 10

then it is algebra from here on in
simplify the complex fraction to get
\[\frac{\frac{5+5x}{x}}{x+1}=\frac{\frac{5(x+1)}{x}}{x+1}=\frac{5}{x}\]

Answer 11

take the limit by replacing \(x\) by \(-1\) and get \(-5\)
of course the shortcut method is quicker

Answer 12

if you want to compute
\[\frac{\frac{5}{x+h}-\frac{5}{x}}{h}\] we can do that too

Answer 13

I'm doing precal so I haven't been introduced to any shortcuts. I think I understand now. Thank you for your help. I just thought you were supposed to use the derivative equation which is what I did. Can we try it that way

Answer 14

I got to that part, but I don't know what to do from there

Answer 15

it is just more algebra
the numerator is
\[\frac{5x-5(x+h)}{x(x+h)}=\frac{-5h}{x(x+h)}\]

Answer 16

then when you divide by \(h\) the \(h\) cancels and you are left with
\[\frac{-5}{x(x+h)}\]

Answer 17

What happened to the second five?

Answer 18

take the limit as \(h\to 0\) and get \(-\frac{5}{x^2}\) then replace \(x\) by \(-1\) and get \(-5\)

Answer 19

lets go slow

Answer 20

Okay.

Answer 21

this is what you have to compute
\[\lim_{h\to 0}\frac{\frac{5}{x+h}-\frac{5}{x}}{h}\]

Answer 22

now lets ignore the \(h\) in the denominator for a moment, and concentrate only on
\[\frac{5}{x+h}-\frac{5}{x}\]

Answer 23

\[\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}\] so
\[\frac{5}{x+h}-\frac{5}{x}=\frac{5x-5(x+h)}{x(x+h)}\] so far so good?

Answer 24

So you cross multiplied and then multiplied the denominators?

Answer 25

yes

Answer 26

just like with numbers right?
\[\frac{5}{7}-\frac{3}{8}=\frac{5\times 8-7\times 3}{7\times 8}\]

Answer 27

yes

Answer 28

now we can compute
\[\frac{5x-5(x+h)}{x(x+h)}=\frac{5x-5x-5h}{x(x+h)}=\frac{-5h}{x(x+h)}\]

Answer 29

ok to there?

Answer 30

okay. Now I see where the second five went. It got cancelled out. I'm ok

Answer 31

now don't forget there was an \(h\) in the denominator of the original expression
when you divide \(\frac{-5h}{x(x+h)}\) by \(h\) you are left only with \(-\frac{5}{x(x+h)}\)

Answer 32

I got that. what next

Answer 33

@satellite73

Answer 34

So where'd you leave off?
You were able to get a common denominator and combine those two fractions?

\[\large \lim_{h \rightarrow 0} \frac{-5}{x(x+h)}\]

Need help from this step?

Answer 35

Yes!

Answer 36

We can from this point plug zero directly in for h, since it no longer causes a problem for us.

Answer 37

and then put -1 in for x^2?

Answer 38

So we started with \(\large f(x)=\dfrac{5}{x}\)

By using the limit definition of the derivative we were able to determine that \(\large f'(x)=-\dfrac{5}{x^2}\)

And now they're asking for \(\large f'(-1)\).

Yes very good :) plug away!

Answer 39

Thank you so much! Would you like to help me with another problem? It involves finding a limit algebraically.

Answer 40

Close this thread, it's gotten rather long and messy.
Open a new one with your new question.

Yah I'll come take a look :)

Answer 41

thank you!