Question

Find the derivative of f(x) = -6/x at x = 12.

Answer 1

@phi

Answer 2

first set up the expression (f(x+h) - f(x)) / h

Answer 3

f'(x)=6/x^2
f'(12)=6/(12^2)

Answer 4

\[f(x)=-\frac{ 6 }{ x }\] at x=12

Answer 5

How would I change it to that?

Answer 6

first , the x=12 is used at the last step.

second, f(x+h) means replace x in your function definition with x+h

Answer 7

fill in what you can in:
\[ \lim_{h \rightarrow 0} \frac{ f(x+h) - f(x)}{h} \]

Answer 8

I understood you to ask for the derivative of f(x) = -6/x = -6x^(-1).
There's a differentiation rule for powers of x that gives (d/dx)x^(-1) = -1x(^-2). This is what I used.
(-6)(-1x^(-2)) = 6/x^2
Or, of course you can take phi's approach which is lower-level and always works.

Answer 9

@BillBell yes, @This_Is_Batman is using the fundamental def of the derivative for his problems.

Answer 10

for this problem, it might be less confusing to write it as
\[ \lim_{h \rightarrow 0} \frac{1}{h} ( f(x+h) - f(x) )
\]
It's the same thing, but less confusing (maybe?)

Answer 11

so far you have
\[ f(x)=-\frac{ 6 }{ x } \]
you also need f(x+h)

Answer 12

Would I do -6/x + -6/h?

Answer 13

Sorry @phi , I was starving to death..

Answer 14

are you asking what is f(x+h) ?

write down f(x)
everywhere you see x, erase it and put in (x+h)

Answer 15

That's what I mean. I have -6/x, so wouldn't I do -6/x+h, which is the same as -6/x -6/h?

Answer 16

you can't split -6/(x+h) (you might be thinking of (x+h)/6 = x/6 + h/6 )

Answer 17

Oh, my bad. :p

Answer 18

@phi: ok, understood

Answer 19

So I just have to figure out -(6/(x+h))-(-(6/x))?

Answer 20

try to be complete
\[ \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{-6}{x+h} - \frac{-6}{x} \right) \]

to add/subtract fractions you need a common denominator.

Answer 21

So now it's:
\[\lim_{h \rightarrow 0}(\frac{ -12h }{ x+h })\]

Answer 22

how did you get that ?
I would multiply the first fraction by x/x
and the second fraction by (x+h)/(x+h)

can you do that ?

Answer 23

Wow, I am retarded. I see what we are doing. It should be:
\[\frac{ -12x - 6h }{ x^2 +h}\]

Answer 24

x+h and x are not the same number (so not the common denominator)

we don't have a way to "add h" to the second x to make it x+h (while keeping the fraction unchanged)
all we know how to do is multiply: x(x+h) is the common denominator

Answer 25

I meant xh, not just h on the denominator.

Answer 26

The derivative of f is the function whose value at x is the limit

Answer 27

closing in on it. but you have a - (-6) so things cancel
also x(x+h) is not x^2 + h

Answer 28

Oh, so just -6h/(x^2+xh)...

Answer 29

Which would simplify down to -6/(x^2+x), right?

Answer 30

double check your algebra. I get +6h up top

Answer 31

Yup, my bad. 6/(x^2+x)

Answer 32

I have to get rid of the x...

Answer 33

try be be more careful
\[ \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{-6}{x+h} - \frac{-6}{x} \right) \\
\lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{6h}{x^2+xh)} \right)
\]

Answer 34

notice that 1/h times h cancel and you get
\[ \lim_{h \rightarrow 0} \left( \frac{6}{x^2+xh} \right) \]

now "take the limit"

Answer 35

I thought the h on the bottom would go away?

Answer 36

Do you see what I posted ? after you put the fractions over a common denominator you get
\[ \lim_{h \rightarrow 0} \frac{1}{h} \left( \frac{6h}{x^2+xh)} \right) \]
at this point we cannot (yet) let h=0 because h/h would be undefined
the next step is simplify h/h to get
\[ \lim_{h \rightarrow 0} \left( \frac{6}{x^2+xh)} \right) \]
now let h approach 0. the xh "goes away"

Answer 37

and the answer is 6/x^2

Answer 38

Oh, so it goes down to 6/x^2. Which would be 1/24.