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 October 25th, 2014, 09:04 PM #1 Newbie   Joined: Jul 2014 From: Brazil Posts: 8 Thanks: 0 How to find f(x)=1/x using the fundamental theorem of calculus? I know how to find it using the short-cut rule, which is basically $\displaystyle f(x)=1/x$ =$\displaystyle x^-1$ =$\displaystyle -x^-2$ But the question is asking me to find the derivative using the fundamental theorem of calculus and I seriously have no idea how and I'm honestly really frustrated because I've been stuck on this for two days now. If you have any idea on how to solve it using the theorem, then I'd really appreciate the help ;-; The formula is supposed to be f(x) = lim h approaches 0 $\displaystyle f(x+h) - f(x)/h$ Last edited by Anon321; October 25th, 2014 at 09:13 PM.
 October 25th, 2014, 09:31 PM #2 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry $\displaystyle f(x)=\frac{1}{x}$ $\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ $\displaystyle =\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$ $\displaystyle =\lim_{h\to0}\frac{\frac{x}{x(x+h)}-\frac{x+h}{x(x+h)}}{h}$ $\displaystyle =\lim_{h\to0}\frac{x-(x+h)}{xh(x+h)}$ $\displaystyle =\lim_{h\to0}\frac{-\cancel{h}}{x\cancel{h}(x+h)}$ $\displaystyle =\lim_{h\to0}\frac{-1}{x(x+h)}$ Substitute h = 0, and you will get $\displaystyle -\frac{1}{x^2}$ Thanks from MarkFL, topsquark, v8archie and 2 others Last edited by Monox D. I-Fly; October 25th, 2014 at 09:48 PM.
October 25th, 2014, 09:48 PM   #3
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Quote:
 Originally Posted by Anon321 =$\displaystyle x^-1$ =$\displaystyle -x^-2$
A quick LaTeX tip: When you have more than one character in the exponent, you need to include all symbols within {.}. For example:
Your code: x^-1 vs. x^{-1} gives

$\displaystyle x^-1$ vs. $\displaystyle x^{-1}$

-Dan

 October 30th, 2014, 08:32 AM #4 Newbie   Joined: Oct 2014 From: earth Posts: 3 Thanks: 0 Last edited by skipjack; November 1st, 2014 at 09:40 AM.
October 30th, 2014, 01:12 PM   #5
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Quote:
 Originally Posted by Anon321 & my edit $\displaystyle f(x)=1/x$ = $\displaystyle \ x^{-1} \ \ \ \ \ \ \ \$(exponent edited) =$\displaystyle \ -x^{-2} \ \ \ \$(exponent edited) This is not true. The function is not equal to its derivative. $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$Don't put a chain of equal signs there. The formula is supposed to be f(x) = lim h approaches 0 $\displaystyle \ \ [f(x+h) - f(x)]/h \ \ \ \$You must have grouping symbols around the numerator. $\displaystyle \ \$ I put brackets in as an example.
.

October 30th, 2014, 06:09 PM   #6
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Quote:
 Originally Posted by Anon321 But the question is asking me to find the derivative using the fundamental theorem of calculus
Note that that $$\frac{\mathbb d x}{\mathbb d y} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$is the definition of the derivative of $f(x)$ at $x$, not the Fundamental Theorem of Calculus.

Last edited by skipjack; November 1st, 2014 at 09:42 AM.

 October 30th, 2014, 11:12 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond Once again, please remain on topic.
October 30th, 2014, 11:14 PM   #8
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Quote:
 Originally Posted by greg1313 Once again, please remain on topic.
I just reported his post for being combative and argumentative.

Last edited by skipjack; November 1st, 2014 at 09:45 AM.

 October 31st, 2014, 06:12 AM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra At the risk of straying further off topic, I would reply to your combative and argumentative private message, but you don't see fit to allow people to send you any private messages so I'll just have to ignore it. [Note by moderator: I've removed the off-topic content with v8archie's consent.] Last edited by skipjack; November 1st, 2014 at 09:51 AM.

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