Question

Find the derivative of f(x) = 1/ x+9 by using the difference quotient.

Answer 1

(x+9)^2?

Answer 2

\[-h / (x+h+9)(x+9)\]

Answer 3

am I right or can I go further

Answer 4

I think you're missing factor of h....? That h in the num'r should cancel with h in the den'r of the original DQ.

Answer 5

\[ \dfrac{ \dfrac{ 1 }{ x+h+9 } - \dfrac{ 1 }{ x+9 }}{ h }\]
\(= \dfrac{ \dfrac{ x+9-(x+h+9) }{ (x+h+9)(x+9) } }{ h }\)

\(= \dfrac{ -h }{ (x+h+9)(x+9) } \cdot \dfrac{1}{ h }\)

Answer 6

So the -h and the h cancel, leaving a -1 in the num'r..... then take the limit.

Answer 7

wow I forgot it pretty much

Answer 8

@asdafogbucket I'm not sure how you got that... that isn't the correct DQ for this problem. I thought at first that you did the DQ for f(x) = 1/ x+9, which IS what was written (he meant f(x) = 1/ (x+9), it seems)... but yours wouldn't be correct for f(x) = 1/ x+9, either.

Answer 9

so m'(x) = \[-1 / (x+9)^2\]

Answer 10

Is the function was, in fact, f(x)=(1/x) + 9, then the DQ would be:

\[\Large \dfrac{ \dfrac{ 1 }{ x+h } + 9 - (\dfrac{ 1 }{ x } + 9)}{ h }\]

the "x+h" must be together in that num'r, and the parentheses around the "-f(x)" part are what I like to call, "NON-optional parentheses!" :)

Answer 11

Yes, exactly @smallmelo - that's it! :)

Answer 12

thanks!

Answer 13

no problem,happy to help. :)