Author Topic: help (Read 6481 times) Tweet Share

TrueTears · « **Reply #15 on:** July 04, 2009, 10:25:00 pm »

I assume you can apply quotient rule and differentiate.

$\frac{d}{dx}\frac{sinx+cosx}{sinx-cosx} = \frac{-2}{(cosx-sinx)^2} = \frac{-2}{1-2sinxcosx} = \frac{2}{sin(2x)-1}$

[I think you mean denominator is sin(2x)-1]

TrueTears · « **Reply #16 on:** July 04, 2009, 10:50:36 pm »

Quote from: Mao on July 04, 2009, 09:41:19 pm

No need to do that

$\frac{\tan(x) +1 }{\tan(x) - 1} = \frac{\tan(x) - 1 + 2}{\tan(x) - 1} = \frac{\tan(x) - 1}{\tan(x) - 1} + \frac{2}{\tan(x)-1}$

true.

khalil · « **Reply #17 on:** July 04, 2009, 11:06:09 pm »

2sinxcosx=sin2x? Do you have proof? I never knew this!

And thanks for your help so far!

Damo17 · « **Reply #18 on:** July 04, 2009, 11:11:00 pm »

Quote from: khalil on July 04, 2009, 11:06:09 pm

2sinxcosx=sin2x? Do you have proof? I never knew this!

And thanks for your help so far!

Double angle formulas are not in the methods course, only in spesh so that is why you have not seen it before.

TrueTears · « **Reply #19 on:** July 04, 2009, 11:12:16 pm »

True they're not in the course but they're useful to know.

You won't be penalized using them and they only make things easier.

Show you proof in a sec

TrueTears · « **Reply #20 on:** July 05, 2009, 12:10:40 am »

Check other thread, I've attached some good shizz if you're interested.

khalil · « **Reply #21 on:** July 05, 2009, 01:15:51 pm »

Which one?

khalil · « **Reply #22 on:** July 05, 2009, 01:49:59 pm »

Find local maximum and minimum for f(x)= 2cos(x) + 2sin(x)cos(x)

TrueTears · « **Reply #23 on:** July 05, 2009, 03:17:25 pm »

Quote from: khalil on July 05, 2009, 01:15:51 pm

Which one?

http://vcenotes.com/forum/index.php/topic,14951.0.html

TrueTears · « **Reply #24 on:** July 05, 2009, 03:26:02 pm »

Quote from: khalil on July 05, 2009, 01:49:59 pm

Find local maximum and minimum for f(x)= 2cos(x) + 2sin(x)cos(x)

$\frac{d}{dx}(2cos(x) + 2sin(x)cos(x)) = 2cos^2x - 2sin^2x - 2sinx = 2cos(2x) - 2sinx = 2(1-2sin^2x)-2sinx = -4sin^2x - 2sinx + 2$

We require $\frac{dy}{dx} = 0$

$-4sin^2x - 2sinx + 2 = 0$

Let $a = sinx$

$-4a^2-2a+2=0$

$a = -1$ or $\frac{1}{2}$

Hence $sinx = -1$ or $sinx = \frac{1}{2}$

Solve for the required domain or just give general solution if no domain is specified.

khalil · « **Reply #25 on:** July 06, 2009, 12:29:00 pm »

Thanks again.

What about this one?

If x= 2e^(-2t)+3e^(-t) show that ((d^2)x)/(d(t^2)) + 3(dx/dt) + 2x = 0

d0minicz · « **Reply #26 on:** July 06, 2009, 12:34:00 pm »

Is the equation $x = 2e^{-2t} + 3e^{-t} + 3e^{-t}$ ?

khalil · « **Reply #27 on:** July 06, 2009, 12:53:03 pm »

no, sorry, just 1 3e^(-t)

cobby · « **Reply #28 on:** July 06, 2009, 01:08:43 pm »

khalil · « **Reply #29 on:** July 06, 2009, 01:35:16 pm »

How did u get the solution from the third line?

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