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Differentiation Question

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Chingoo:
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 =/

Dibss:
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)

(:

Arthur Bon Zavi:

--- Quote from: Whisper 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)

(:

--- End quote ---

 + rep. ;)

Dibss:
Thanks. (:

Chingoo:
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.

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