Author Topic: Cobby's Methods Questions (Read 40908 times) Tweet Share

cobby · « **Reply #76 on:** April 07, 2009, 08:28:03 pm »

Find the value of x
$x \epsilon [-\pi,\pi]$
$\cos x = -\sin \frac{\pi}{6}$

Thanks guys

TrueTears · « **Reply #77 on:** April 07, 2009, 08:28:58 pm »

$cos(x) = - sin(\frac{\pi}{6})$

$sin(\frac{\pi}{2}-x) = sin(\frac{7\pi}{6}) , sin(\frac{11\pi}{6})$

$\frac{\pi}{2} - x = \frac{7\pi}{6} , \frac{11\pi}{6}$

$x = -\frac{2}{3}\pi , -\frac{4}{3}\pi$

but domain is $\frac{\pi}{2} < x < \pi$

so $-\frac{4}{3}\pi = \frac{2\pi}{3}$ ( $-\frac{2}{3}\pi$ is in the 4th quadrant, so can not be used in this domain)

therefore $x = \frac{2\pi}{3}$

Mao · « **Reply #78 on:** April 07, 2009, 10:31:31 pm »

that's a rather round-about way of doing this question.

$\cos (x) = -\sin \frac{\pi}{6}$

$\implies \cos(x) = -\frac{1}{2}$

$\implies x = \pm \frac{2\pi}{3}$

TrueTears · « **Reply #79 on:** April 07, 2009, 10:34:00 pm »

Quote from: Mao on April 07, 2009, 10:31:31 pm

that's a rather round-about way of doing this question.

$\cos (x) = -\sin \frac{\pi}{6}$

$\implies \cos(x) = -\frac{1}{2}$

$\implies x = \pm \frac{2\pi}{3}$

book said not to evaluate $sin(\frac{pi}{6})$ lols

Mao · « **Reply #80 on:** April 07, 2009, 10:34:55 pm »

oh, I see

my bad

kamil9876 · « **Reply #81 on:** April 07, 2009, 10:50:39 pm »

Would've been better if the angle wasn't exact, then you would be forced to take the 'round-about'

cobby · « **Reply #82 on:** April 30, 2009, 11:56:44 am »

Hey guys another question please..

Differentiate using Product Rule
$x^3 (3x^2+2x+1)^{-1}$

Thanks

TrueTears · « **Reply #83 on:** April 30, 2009, 12:28:56 pm »

$3x^2(3x^2+2x+1)^{-1} + x^3(-1)(3x^2+2x+1)^{-2}(6x+2)$

I assume you can simplify from there?

TrueTears · « **Reply #84 on:** May 04, 2009, 03:30:52 pm »

(as request)

continuing from last line of working.

$\frac{3x^2}{3x^2+2x+1} - \frac{x^3(6x+2)}{(3x^2+2x+1)^2}$

$= \frac{3x^2(3x^2+2x+1)}{(3x^2+2x+1)^2} - \frac{x^3(6x+2)}{(3x^2+2x+1)^2}$

$= \frac{3x^2(3x^2+2x+1) - x^3(6x+2)}{(3x^2+2x+1)^2}$

$= \frac{3x^4 + 4x^3 + 3x^2}{(3x^2+2x+1)^2}$

cobby · « **Reply #85 on:** May 08, 2009, 08:33:16 pm »

Hey guys,

Is it possible to find the derivative of $|x-1|-1$ by hand?

I typed it in on my calculator and it came up with $sign(x-1)$

Mao · « **Reply #86 on:** May 08, 2009, 08:59:47 pm »

Think of the modulus function as a hybrid function of positive and negative parts

$|x-1|-1 = \begin{cases} -(x-1)-1 & x\le 1 \\ (x-1)-1 & x > 1 \end{cases}$

simplifying and differentiating both side gives

$\frac{d}{dx}(|x-1|-1) = \begin{cases} -1 & x < 1 \\ 1 & x > 1 \end{cases}$

cobby · « **Reply #87 on:** May 10, 2009, 12:31:08 pm »

Hey guys,
Struggling with this question..

if $f(x) = (x+1) (\sqrt {x^2+1})$

Find $f'(x)$

Thanks

dekoyl · « **Reply #88 on:** May 10, 2009, 12:49:29 pm »

Use the product rule:
$f'(x)=(x+1)\frac{2x}{2\sqrt{x^2+1}} + \sqrt{x^2+1}$
$=\frac{x(x+1)}{\sqrt{x^2+1}} + \sqrt{x^2+1}$
I think that's all you need to do.

cobby · « **Reply #89 on:** May 10, 2009, 06:53:14 pm »

Thanks dekoyl...i got that answer too..

but the book says..
$f'(x)= \frac {2x^2+x+1}{\sqrt{x^2+1}}$

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