anonymous
  • anonymous
Find the derivative of f(x) = 6/x at x = -2.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
solomon help haha
SolomonZelman
  • SolomonZelman
Ok, lets re-write the f(x) \(\large\color{black}{ \displaystyle f(x)=6(x)^{-1} }\)
SolomonZelman
  • SolomonZelman
Apply the power rule. Can you do that?

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SolomonZelman
  • SolomonZelman
(YOu are to find the derivative, and then plug in x=-2 into the derivative)
anonymous
  • anonymous
-3
SolomonZelman
  • SolomonZelman
is that your fnal answer?
SolomonZelman
  • SolomonZelman
if so, then you are not correct....
SolomonZelman
  • SolomonZelman
Did you find the \(f'(x)\) /?
SolomonZelman
  • SolomonZelman
Oh, what I mean by the power rule is: \(\large\color{black}{ \displaystyle \frac{d }{dx}x^n=nx^{n-1} }\) where d/dx is jst a notation for taking the derivative. ------------------------------------------------ But I guess you are doing by the first principles...
anonymous
  • anonymous
'm not really sure what you mean by power rule I have a formula for difference quotient f(h-1)-f(1)/h and I ended up with ([6/h-1]-6)/h
anonymous
  • anonymous
never heard of it this is precalc
SolomonZelman
  • SolomonZelman
yes, you are applying the following: \(\large\color{black}{ \displaystyle \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} }\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} }\)
SolomonZelman
  • SolomonZelman
YOu can use the power rule I posted, to at least check the work, but for now I guess we need this: \(\large\color{black}{ \displaystyle \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} }\)
SolomonZelman
  • SolomonZelman
Or, if you want to find \(f'(a)\) direclty: \(\large\color{black}{ \displaystyle f'(a)= \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} }\)
anonymous
  • anonymous
(6/x+h)-(6/x)/h
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6}{x+h}-\dfrac{6}{x} }{h} }\)
SolomonZelman
  • SolomonZelman
that is right.
SolomonZelman
  • SolomonZelman
now, find the common denominator betwen 6/(x+h) and 6/x and subtract.
anonymous
  • anonymous
so would I multiply one side by x and the other by x+h
SolomonZelman
  • SolomonZelman
yes, fraction#1 by x on top and bottom, and fraction#2 (x+h) on top and bottom
anonymous
  • anonymous
I got -6
anonymous
  • anonymous
pretty sure that's it thanks
SolomonZelman
  • SolomonZelman
no it is not it
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6}{x+h}-\dfrac{6}{x} }{h} }\) \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6x}{x(x+h)}-\dfrac{6(x+h)}{x(x+h)} }{h} }\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{6x-6(x+h)}{x(x+h)} }{h} }\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{-6h}{x(x+h)} }{h} }\)
anonymous
  • anonymous
yep that's what I got just forgot about the denomenator under the -6h
SolomonZelman
  • SolomonZelman
then divide top and bottom by h.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}\frac{\dfrac{-6h}{x(x+h)}\color{red}{\div h} }{h \color{red}{\div h}} }\)
SolomonZelman
  • SolomonZelman
then you get: \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}{~} \frac{-6}{x(x+h)} }\)
anonymous
  • anonymous
-6/x(x+h)
SolomonZelman
  • SolomonZelman
yes, but you are leaving out that important limit.
anonymous
  • anonymous
so I just put -2 in for x
SolomonZelman
  • SolomonZelman
that limit that h=0, is an important component. So that when you simplify the expression, you then plug in h=0 (if you don;t get any undefined results for that)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}{~} \frac{-6}{x(x+h)} =\frac{-6}{x(x+0)}=\frac{-6}{x^2} }\)
SolomonZelman
  • SolomonZelman
see what is that limit for? (it is a notation for the fact that h is 0)
anonymous
  • anonymous
- 3/2
SolomonZelman
  • SolomonZelman
yes, that is right:
anonymous
  • anonymous
hey, thanks for the patience I'm kind of slow
SolomonZelman
  • SolomonZelman
Yes, don't forget that limit h->0 notation. it is important.
SolomonZelman
  • SolomonZelman
So just an addition that in general: \(\large\color{black}{ \displaystyle f'(x)= \lim_{h \rightarrow 0}{~} \frac{f(x+h)-f(x)}{h} }\) (Derivative a function f(x).) \(\large\color{black}{ \displaystyle f'(x)= \lim_{x \rightarrow a}{~} \frac{f(x)-f(a)}{x-a} }\) (Derivative a function f(x) evaluated at x=a.)
SolomonZelman
  • SolomonZelman
good luck

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