Author Topic: double derivative questions help (Read 1404 times) Tweet Share

yokomodo · « **on:** July 02, 2008, 01:21:20 am »

hi vn, first post.

i need some help with these questions, 3,5,7, (answers are at the 2nd half of the pic.)

i dont seem to get the right answer.

:uglystupid2: :idiot2: :2funny:
thanks

dcc · « **Reply #1 on:** July 02, 2008, 08:14:43 am »

Q3:

$\mbox{Let u }= 2x,\; y = \tan^{-1} u$

$\therefore \dfrac{dy}{du} = \dfrac{1}{1 + u^2} \implies \dfrac{dy}{dx} = 2 \cdot \dfrac{1}{1 + (2x)^2} = \dfrac{2}{1 + 4x^2}$

Then finding the derivative of this:

$\dfrac{d^2y}{dx^2} = \dfrac{-2(8x)}{(1 + 4x^2)^2}= - \dfrac{16x}{(1 + 4x^2)^2}$

So the answer given is wrong.

Q5:
$y = e^{-mx} \implies \dfrac{dy}{dx} = -m \cdot e^{-mx} \implies \dfrac{d^2y}{dx^2} = m^2 \cdot e^{-mx}$

Given that $y = \dfrac{d^2y}{dx^2} \mbox{ } \forall x \in R$

$m^2 \cdot e^{-mx} = e^{-mx} \implies m = \pm 1\mbox{ (as $e^{-mx}$ $>$ 0 always)}$

kj_ · « **Reply #2 on:** July 02, 2008, 08:16:17 am »

Q7:

$y=e^x$
$y'=e^x$
$y''=e^x$
(yes, they're all the same - you should know this, yeah?

$therefore, <br />\frac{1}{e^x}*e^x - \frac{2x}{e^x}*e^x$
$=1-2x$

AppleXY · « **Reply #3 on:** July 02, 2008, 08:43:06 am »

Q6.

$y = \sin(x^2)$

$\mbox{Let u = x^2}$

$\therefore y = \sin(u)$

$\mbox{using the chain rule}$

$\frac{dy}{dx} = \frac{du}{dx} \cdot \frac{dy}{du}$

$\frac{dy}{dx} = 2x \cdot cos(u)$

$\therefore \frac{dy}{dx} = 2x\cos(x^2)$

for $\frac{d^{2}x}{dy^2}$ use the product rule.

$\mbox{Let u = 2x and v = cos(x^2)}$

$\therefore \frac{dy}{dx} = u \cdot v$

$\frac{d^{2}x}{dy^2} = v \cdot \frac{du}{dx} + u \cdot \frac{dv}{dx}$

$\therefore \frac{d^{2}x}{dy^2} = 2cos(x^2) - 4x^2\sin(x^2)$

OHSHI- YOU SAID 3,5 and 7. F** oh well what a waste of coding time fk.

dcc · « **Reply #4 on:** July 02, 2008, 09:00:33 am »

lol prav

beezy4eva · « **Reply #5 on:** July 02, 2008, 09:30:04 am »

3. y=tan-1(2x)
.5 tan(y)= x
dx/dy= .5 sec²(y) -using trigonometric identities
dx/dy= .5 (tan² (y) + 1)
dy/dx= 1/.5(tan²(y) +1) -sub in y=tan-1(2x)
dy/dx= 2/((4x² +1))
d²y/dx²= -2(4x²+1)-².8x
d²y/dx²= -16x/(4x²+1)²

me thinks i've gone wrong somewhere....

Mao · « **Reply #6 on:** July 02, 2008, 10:12:36 am »

it's right =]

antidifferentiating the textbook answer gives us:

$- 2\cdot \tan^{-1}(2x) - 2x\cdot ln(4x^2 + 1) + C\cdot x + D$

yokomodo · « **Reply #7 on:** July 02, 2008, 07:59:29 pm »

thanks for the help/replies guys, really appreciated

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Author Topic: double derivative questions help (Read 1404 times) Tweet Share

yokomodo

double derivative questions help

dcc

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AppleXY

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dcc

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beezy4eva

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Mao

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yokomodo

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