Author Topic: Differentiation (Read 6860 times) Tweet Share

dcc · « **Reply #30 on:** June 16, 2009, 08:13:33 pm »

Quote from: d0minicz on June 16, 2009, 07:57:11 pm

$\int x \sqrt {4-x}\: dx$
need to check if the answer is right ta =]

Let $u = 4 - x \implies \frac{du}{dx} = -1$

Therefore our integral becomes $I = \int -\left(4 - u\right) \sqrt{u}\: du = - \int \left( 4 u^{\frac{1}{2}} - u^{\frac{3}{2}} \right)\: du$

$I = - \left( 4u^{\frac{3}{2}} \cdot \frac{2}{3} - u^{\frac{5}{2}} \cdot \frac{2}{5} \right) + C$ , $C \in \mathbb{R}$ .

$\implies \int x \sqrt{4 - x}\: dx = \left(4 - x\right)^{\frac{3}{2}}\left( \frac{2}{5} \left(4 - x\right) - \frac{8}{3} \right) + C$

This can be simplified further, but I'm a lazy guy, so

d0minicz · « **Reply #31 on:** June 16, 2009, 10:08:43 pm »

$\int^m_0 tanx sec^2x dx = \frac {3}{2}$ when $m \in (0,\frac {\pi}{2})$ . The value of m is ?
thanks =]

ilovevce · « **Reply #32 on:** June 16, 2009, 10:40:33 pm »

Quote from: d0minicz on June 16, 2009, 10:08:43 pm

$\int^m_0 tanx sec^2x dx = \frac {3}{2}$ when $m \in (0,\frac {\pi}{2})$ . The value of m is ?
thanks =]

Let u = tan(x), then we have $\frac{du}{dx} = sec^2x$

Also, when x=0, u = tan(0) = 0
when x = m, u = tan(m)

So now we have $\int_0^{tan(m)} u du = \frac {3}{2}$

$\implies \frac{tan^2(m)}{2} = \frac{3}{2}$

$\therefore tan(m) = \pm\sqrt{3}$ where $m \in (0,\frac {\pi}{2})$

$\therefore m = \frac{\pi}{3}$

d0minicz · « **Reply #33 on:** June 17, 2009, 06:50:45 pm »

Find the second derivative of :
$f(x) = sin(e^x)$
and $f(x)= xsin^{-1}x$
thank you

TrueTears · « **Reply #34 on:** June 17, 2009, 06:53:02 pm »

$f'(x) = sin^{-1}(x) + x(\frac{1}{\sqrt{1-x^2}})$

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

Then simplify.

d0minicz · « **Reply #35 on:** October 25, 2009, 12:59:03 pm »

Question: $\frac {d}{dx} (cos^{-1} (\sqrt {3x}))$

i cant seem to get their form ... thanks.

kamil9876 · « **Reply #36 on:** October 25, 2009, 01:03:18 pm »

looks similair to: http://vcenotes.com/forum/index.php/topic,11962.msg184369.html#msg184369

d0minicz · « **Reply #37 on:** October 25, 2009, 01:07:29 pm »

ah yeah i got that form

their answer: $\frac {-\sqrt{3}}{2\sqrt{x-3x^2}}}$

humph · « **Reply #38 on:** October 25, 2009, 01:16:56 pm »

Quote from: d0minicz on October 25, 2009, 01:07:29 pm

ah yeah i got that form

their answer: $\frac {-\sqrt{3}}{2\sqrt{x-3x^2}}}$

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

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Author Topic: Differentiation (Read 6860 times) Tweet Share

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