hopelynnwelch
Junior Member
- Joined
- Jan 16, 2015
- Messages
- 60
Okay, so I have this problem and I'm not allowed to use any product rules or anything like that because all that is in the next chapter. So I got some help from a tutor to do this problem and here is what I have. Does this look right to you guys?
\(\displaystyle f(x)\, =\, 2\sqrt{x\, +\, 1\,}\, +\, \dfrac{3}{x\, +\, 2}\, -\, \dfrac{1}{3}\, x^2\)
\(\displaystyle f'(x)\, =\, \displaystyle{ \lim_{h\, \rightarrow\, 0} \,}\)\(\displaystyle \, \dfrac{2\sqrt{(x\, +\, h)\, +\, 1\,}\, -\, 2\sqrt{x\, +\, 1\,}\, +\, \dfrac{3}{(x\, +\, h)\, +\, 2}\, -\, \dfrac{3}{x\, +\, 2}\, -\, \dfrac{(x\, +\, h)^2}{3}\, +\, \dfrac{x^2}{3}}{h}\)
Taking just the first two terms in the numerator above, and working just with them:
\(\displaystyle \dfrac{\left(2\sqrt{x\, +\, h\, +\, 1\,}\, -\, 2\sqrt{x\, +\, 1\,}\right)\, \left(2\sqrt{x\, +\, h\, +\, 1\,}\, +\, 2\sqrt{x\, +\, 1\,}\right)}{h\, \left(2\sqrt{x\, +\, h\, +\, 1\,}\, +\, 2\sqrt{x\, +\, 1\,}\right)}\, =\, \dfrac{4(x\, +\, h\, +\, 1)\, -\, 4(x\, +\, 1)}{2h\, \left(\sqrt{x\, +\, h\, +\, 1\,}\, +\, \sqrt{x\, +\, 1\,}\right)}\)
\(\displaystyle =\, \dfrac{1}{2\,\left(\sqrt{x\, +\, h\, +\, 1\,}\, +\,\sqrt{x\, +\, 1\,}\right)}\, =\, \color{blue}{ \dfrac{1}{4\, \sqrt{x\, +\, 1\,}} }\)
Taking the middle two terms in the second line above, and then multiplying through by \(\displaystyle \, \dfrac{x\, +\, h\, +\, 2}{x\, +\, h\, +\, 2}:\)
\(\displaystyle \dfrac{\left(\dfrac{x\, +\, h\, +\, 2}{x\, +\, h\, +\, 2}\right)\, \left(\dfrac{3}{x\, +\, h\, +\, 2}\right)\, -\, \left(\dfrac{3}{x\, +\, 2}\right)\, \left(\dfrac{x\, +\, h\, +\, 2}{x\, +\, h\, +\, 2}\right)}{h}\)
\(\displaystyle =\, \dfrac{3(x\, +\, 2)\, -\, 3(x\, +\, h\, +\, 2)}{(x\, +\, h\, +\, 2)\, (x\, +\, 2)}\, \cdot\, \dfrac{1}{h}\, =\, \dfrac{3x\, +\, 6\, -\, 3x\, -\, 3h\, -\, 6}{x^2\, +\, 4x\, +\, 4\, +\, xh\, +\, 2h}\, \cdot\, \dfrac{1}{h}\)
\(\displaystyle =\, \dfrac{-3}{x^2\, +\, 4x\, +\, 4}\, =\, \color{red}{ \dfrac{-3}{(x\, +\,2)^2} }\)
And now doing just the last two terms:
\(\displaystyle \dfrac{-\left[ \dfrac{(x\, +\, h)^2}{3}\, -\, \dfrac{x^2}{3}\right]}{h}\, =\, \dfrac{x^2\, -\, 2xh\, +\, h^2\, -\, x^2}{h}\)
\(\displaystyle =\, \dfrac{-h^2\, -\, 2xh}{3h}\, =\, \dfrac{-h\, -\, 2x}{3}\, =\, \color{green}{ \dfrac{-2x}{3} }\)
Then putting it all together:
\(\displaystyle f'(x)\, =\, \displaystyle{ \lim_{h\, \rightarrow\, 0} \, }\) \(\displaystyle \, \color{blue}{ \dfrac{1}{4\, \sqrt{x\, +\, 1\,}} }\, +\, \color{red}{ \dfrac{-3}{(x\, +\,2)^2} }\,-\, \color{green}{ \dfrac{2x}{3} }\)
Sorry if the work is a little cluttered.
\(\displaystyle f(x)\, =\, 2\sqrt{x\, +\, 1\,}\, +\, \dfrac{3}{x\, +\, 2}\, -\, \dfrac{1}{3}\, x^2\)
\(\displaystyle f'(x)\, =\, \displaystyle{ \lim_{h\, \rightarrow\, 0} \,}\)\(\displaystyle \, \dfrac{2\sqrt{(x\, +\, h)\, +\, 1\,}\, -\, 2\sqrt{x\, +\, 1\,}\, +\, \dfrac{3}{(x\, +\, h)\, +\, 2}\, -\, \dfrac{3}{x\, +\, 2}\, -\, \dfrac{(x\, +\, h)^2}{3}\, +\, \dfrac{x^2}{3}}{h}\)
Taking just the first two terms in the numerator above, and working just with them:
\(\displaystyle \dfrac{\left(2\sqrt{x\, +\, h\, +\, 1\,}\, -\, 2\sqrt{x\, +\, 1\,}\right)\, \left(2\sqrt{x\, +\, h\, +\, 1\,}\, +\, 2\sqrt{x\, +\, 1\,}\right)}{h\, \left(2\sqrt{x\, +\, h\, +\, 1\,}\, +\, 2\sqrt{x\, +\, 1\,}\right)}\, =\, \dfrac{4(x\, +\, h\, +\, 1)\, -\, 4(x\, +\, 1)}{2h\, \left(\sqrt{x\, +\, h\, +\, 1\,}\, +\, \sqrt{x\, +\, 1\,}\right)}\)
\(\displaystyle =\, \dfrac{1}{2\,\left(\sqrt{x\, +\, h\, +\, 1\,}\, +\,\sqrt{x\, +\, 1\,}\right)}\, =\, \color{blue}{ \dfrac{1}{4\, \sqrt{x\, +\, 1\,}} }\)
Taking the middle two terms in the second line above, and then multiplying through by \(\displaystyle \, \dfrac{x\, +\, h\, +\, 2}{x\, +\, h\, +\, 2}:\)
\(\displaystyle \dfrac{\left(\dfrac{x\, +\, h\, +\, 2}{x\, +\, h\, +\, 2}\right)\, \left(\dfrac{3}{x\, +\, h\, +\, 2}\right)\, -\, \left(\dfrac{3}{x\, +\, 2}\right)\, \left(\dfrac{x\, +\, h\, +\, 2}{x\, +\, h\, +\, 2}\right)}{h}\)
\(\displaystyle =\, \dfrac{3(x\, +\, 2)\, -\, 3(x\, +\, h\, +\, 2)}{(x\, +\, h\, +\, 2)\, (x\, +\, 2)}\, \cdot\, \dfrac{1}{h}\, =\, \dfrac{3x\, +\, 6\, -\, 3x\, -\, 3h\, -\, 6}{x^2\, +\, 4x\, +\, 4\, +\, xh\, +\, 2h}\, \cdot\, \dfrac{1}{h}\)
\(\displaystyle =\, \dfrac{-3}{x^2\, +\, 4x\, +\, 4}\, =\, \color{red}{ \dfrac{-3}{(x\, +\,2)^2} }\)
And now doing just the last two terms:
\(\displaystyle \dfrac{-\left[ \dfrac{(x\, +\, h)^2}{3}\, -\, \dfrac{x^2}{3}\right]}{h}\, =\, \dfrac{x^2\, -\, 2xh\, +\, h^2\, -\, x^2}{h}\)
\(\displaystyle =\, \dfrac{-h^2\, -\, 2xh}{3h}\, =\, \dfrac{-h\, -\, 2x}{3}\, =\, \color{green}{ \dfrac{-2x}{3} }\)
Then putting it all together:
\(\displaystyle f'(x)\, =\, \displaystyle{ \lim_{h\, \rightarrow\, 0} \, }\) \(\displaystyle \, \color{blue}{ \dfrac{1}{4\, \sqrt{x\, +\, 1\,}} }\, +\, \color{red}{ \dfrac{-3}{(x\, +\,2)^2} }\,-\, \color{green}{ \dfrac{2x}{3} }\)
Sorry if the work is a little cluttered.
Last edited by a moderator: