Author Topic: Differentiation Question  (Read 3232 times)

Offline Chingoo

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Differentiation Question
« on: December 27, 2010, 02:19:21 pm »
Given that y = (1-sinx)/cosx, show that dy/dx = 1/(1+sinx).

It seems simple but idk why I'm not getting the answer =/
All that is on earth will perish:
But will abide (forever) the Face of thy Lord--full of Majesty, Bounty & Honor.
Then which of the favors of your Lord will ye deny?


Qura'n, Chapter 55: The Beneficent, Verses 26-28

Offline Dibss

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Re: Differentiation Question
« Reply #1 on: December 27, 2010, 09:33:31 pm »
Let S be sinx & C be cosx.
Use the y=u/v => dy/dx = [Udv/dx - Vdu/dx]/V2
Therefore
dy/dx = [(1-S)(-S) - C(0-C)]/C2

[-S + S2 + C2]/C2

Using identity: C2=1-S2

[-S + S2 + 1 - S2]/(1-S2)

(-S+1)/(1-S2))

Use identity: a^2 - b^2 = (a+b) (a-b)

(1-S)/(1+S)(1-S)

Finally dy/dx = 1/(1+sinx)

(:

Offline Arthur Bon Zavi

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Re: Differentiation Question
« Reply #2 on: December 28, 2010, 04:08:37 am »
Let S be sinx & C be cosx.
Use the y=u/v => dy/dx = [Udv/dx - Vdu/dx]/V2
Therefore
dy/dx = [(1-S)(-S) - C(0-C)]/C2

[-S + S2 + C2]/C2

Using identity: C2=1-S2

[-S + S2 + 1 - S2]/(1-S2)

(-S+1)/(1-S2))

Use identity: a^2 - b^2 = (a+b) (a-b)

(1-S)/(1+S)(1-S)

Finally dy/dx = 1/(1+sinx)

(:

 + rep. ;)

Continuous efforts matter more than the outcome.
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Offline Dibss

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Re: Differentiation Question
« Reply #3 on: December 28, 2010, 10:25:57 am »
Thanks. (:

Offline Chingoo

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Re: Differentiation Question
« Reply #4 on: December 28, 2010, 03:40:31 pm »
Thanks, but now I'm confused. Isn't the quotient rule for differentiation
y=u/v => dy/dx = [v.du/dx - u.dv/dx]/v^2
Granted you managed to get the answer, but I'm pretty sure that's the quotient rule.
All that is on earth will perish:
But will abide (forever) the Face of thy Lord--full of Majesty, Bounty & Honor.
Then which of the favors of your Lord will ye deny?


Qura'n, Chapter 55: The Beneficent, Verses 26-28

Offline Dibss

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Re: Differentiation Question
« Reply #5 on: December 28, 2010, 07:04:00 pm »
Sorry, I haven't used the rule for months now - rusty. :S
My bad. So you probably got the answer after using the correct rule right?

Offline Dibss

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Re: Differentiation Question
« Reply #6 on: December 29, 2010, 09:36:12 am »
&& you get the correct answer using the product rule

y=(1-S)(C-1)

dy/dx = (1-S)(S/C2)+(C-1)(0-C)
=S(1-S))/(1-S2) + 1/C (-C)
=S(1-S)/(1+S)(1-S)  - 1
=(S + 1 - S)/(1+S)
=1/(1+S)

Offline Chingoo

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Re: Differentiation Question
« Reply #7 on: December 29, 2010, 07:24:07 pm »
Thanks a lot! =D +rep
All that is on earth will perish:
But will abide (forever) the Face of thy Lord--full of Majesty, Bounty & Honor.
Then which of the favors of your Lord will ye deny?


Qura'n, Chapter 55: The Beneficent, Verses 26-28

Offline Dibss

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Re: Differentiation Question
« Reply #8 on: December 29, 2010, 07:26:27 pm »
No problemm. (:

Offline Tohru Kyo Sohma

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Re: Differentiation Question
« Reply #9 on: December 31, 2010, 04:41:45 pm »
i kind of find maths AS very hard....especially differentiation.....i have a doubt here:
find the coordinates of the point on the curve y=2x^2 -x-1, where the tangent is parallel to the line; y=3x-2?

Offline Dibss

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Re: Differentiation Question
« Reply #10 on: December 31, 2010, 06:59:27 pm »
Practice and you'll get used to it eventually. (:

gradient of line y=3x-2 is 3 b/c in y=mx+c form m is the gradient.
So now we have to find the point on the curve where gradient is 3.
dy/dx= 4x-1
tangent to curve parallel where 4x-1=3
4x=4
x=1
at x=1, y=2-1-1=0
therefore coordinates should be (1,0)
(:

Offline Arthur Bon Zavi

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Re: Differentiation Question
« Reply #11 on: January 01, 2011, 08:00:18 am »
Differentiation is easy and Integration is easier.

Continuous efforts matter more than the outcome.
- NU

Offline Tohru Kyo Sohma

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Re: Differentiation Question
« Reply #12 on: January 01, 2011, 11:38:44 am »
Practice and you'll get used to it eventually. (:

gradient of line y=3x-2 is 3 b/c in y=mx+c form m is the gradient.
So now we have to find the point on the curve where gradient is 3.
dy/dx= 4x-1
tangent to curve parallel where 4x-1=3
4x=4
x=1
at x=1, y=2-1-1=0
therefore coordinates should be (1,0)
(:
thank you so much wispher
i got it now...........;)

Offline Tohru Kyo Sohma

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Re: Differentiation Question
« Reply #13 on: January 01, 2011, 01:00:32 pm »
i have a doubt here;
The production cost per kilogram C(in thousand pounds) when x kilogram of a chemical are made, is given by;
C= 3x+100/x , x>0. find the value for x for which the cost is minimum and the minimum cost??

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Re: Differentiation Question
« Reply #14 on: January 01, 2011, 01:02:31 pm »
Differentiate the equation. Set dy/dx = 0 and solve for x.

Determine if the stationary point is a minimum or maximum. Select the minimum point and plug it into the original equation.

This is the minimum cost.