Author Topic: Lachymm's Methods questions and answer thread (Read 7661 times) Tweet Share

lachymm · « **on:** March 21, 2010, 02:50:41 pm »

hey like some other guys on here i am going to have alot of questions over the next year or so....

I was hoping others would be able to answer questions and others explain how to answer them.

Others can pop in their 2 cents or ask their own questions as they wish.

Alright i will kick it off

$(x^2+1)(\sqrt{2x-4})$ How do i derive fully?

I get upto... $(2x(2x-4)^\frac{1}{2}+(x^2+1)(2x-4)^\frac{-1}{2}$

That latex took me like 15 mins...

the.watchman · « **Reply #1 on:** March 21, 2010, 03:16:50 pm »

Lol, LaTeX isn't that bad...

Anyway:

Product Rule - $\frac{d}{dx}(...) = [x^2+1]' \cdot (2x-4)^{\frac{1}{2}} + (x^2+1) \cdot [(2x-4)^{\frac{1}{2}}]'$

$\frac{d}{dx}(...) = 2x \cdot (2x-4)^{\frac{1}{2}} + (x^2+1) \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2x-4}} \cdot [2x-4]'$

$\frac{d}{dx}(...) = 2x(2x-4)^{\frac{1}{2}} + (x^2+1) \cdot \frac{1}{\sqrt{2x-4}}$

$\frac{d}{dx}(...) = (2x-4)^{\frac{1}{2}}\Big[2x + \frac{x^2+1}{2x-4}\Big]$

m@tty · « **Reply #2 on:** March 21, 2010, 03:19:17 pm »

Quote from: the.watchman on March 21, 2010, 03:16:50 pm

Lol, LaTeX isn't that bad...

.
.
.

$\frac{d}{dx}(...) = 2x \cdot (2x-4)^{\frac{1}{2}} + (x^2+1) \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2x-4} \cdot [2x-4]'$

the.watchman · « **Reply #3 on:** March 21, 2010, 03:19:32 pm »

Already fixed

lachymm · « **Reply #4 on:** March 21, 2010, 03:26:17 pm »

Thanks watchman but i have already gotten upto that part (i said in question)

I dont think i worded it right, i ment how do i factorize fully?

the.watchman · « **Reply #5 on:** March 21, 2010, 03:28:44 pm »

Edited above, how's that

m@tty · « **Reply #6 on:** March 21, 2010, 03:30:51 pm »

Or...

$\frac{d}{dx}(...) = 2x(2x-4)^{\frac{1}{2}} + (x^2+1) \cdot \frac{1}{\sqrt{2x-4}}$

$= \frac{1}{\sqrt{2x-4}}\left(2x(2x-4)+x^2+1\right)$

$= \frac{1}{\sqrt{2x-4}}(5x^2-8x+1)$

$= \frac{5x^2-8x+1}{\sqrt{2x-4}}$

Not saying mine is simpler or anything..

lachymm · « **Reply #7 on:** March 21, 2010, 03:35:39 pm »

i was just posting about how the book says something different to watchmans answer, then i go back to the thread and M@tty has posted it in the form of the book

Thanks to you both. Again i wasted like 10 mins in latex trying to get the dividing and square root signs in the right place...

m@tty · « **Reply #8 on:** March 21, 2010, 03:39:21 pm »

Yeah, he didn't fully simplify it. If he had simplified the inside of the bracket he would arrive at the textbooks solution.

lachymm · « **Reply #9 on:** March 21, 2010, 03:44:52 pm »

Still trying to piece together how you did that matty, so.. $\frac{1}{\sqrt{2x-4}}$ times $2x(2x-4)$ is equal to $2x(2x-4)^\frac{1}{2}$

m@tty · « **Reply #10 on:** March 21, 2010, 03:48:03 pm »

Yeah.

$\frac{1}{\sqrt{2x-4}} \cdot 2x(2x-4)$

$= 2x \cdot \frac{2x-4}{(2x-4)^{\frac{1}{2}}}$

$= 2x \cdot (2x-4)^{1-\frac{1}{2}}$

$= 2x\sqrt{2x-4}$

EDIT: This is only true for $x\ne2$ , this is because in the original expression is undefined for $x=2$ .

lachymm · « **Reply #11 on:** March 21, 2010, 03:54:15 pm »

Another question

Derive $\frac{(x+2)^3}{x^2+1}$

soo.. v= $x^2+1$ v'= $2x$
u= $(x+2)^3$ u'= $3(x+2)^2 (this is after i have done chainrule)$

So $\frac{(x^2+1)(3(x+2)^2)-((x+2)^3)(2x)}{(x^2+1)^2}$

Is this right and if so how do i simplify?

lachymm · « **Reply #12 on:** March 21, 2010, 03:56:04 pm »

sorry for so many brackets

superflya · « **Reply #13 on:** March 21, 2010, 03:58:09 pm »

let $u=x^{2} +1$ .

m@tty · « **Reply #14 on:** March 21, 2010, 04:06:43 pm »

Quote from: lachymm on March 21, 2010, 03:54:15 pm

Another question

Derive $\frac{(x+2)^3}{x^2+1}$

soo.. v= $x^2+1$ v'= $2x$
u= $(x+2)^3$ u'= $3(x+2)^2 (this is after i have done chainrule)$

So $\frac{(x^2+1)(3(x+2)^2)-((x+2)^3)(2x)}{(x^2+1)^2}$

Is this right and if so how do i simplify?

$=\frac{(x+2)^2}{(x^2+1)^2} \cdot \left(3x^2+3-(x+2)(2x)\right)$

$= \frac{(x+2)^2}{(x^2+1)^2} \cdot \left(x^2-4x+3\right)$

$= \frac{(x+2)^2}{(x^2+1)^2} \cdot (x-3)(x-1)$

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