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Math Help - Derivative of Trig Function using quotient rule Help

  1. #1
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    Derivative of Trig Function using quotient rule Help

    Hey guys need some help here. The trig identities are killing me so far this semester. i have to use the quotient rule. Here is the problem and where i have gotten to.

    y= sinx+cosx/cosx
    y'= f'g-fg'/g^2
    = {[(cosx-sinx)(cosx)]-[(sinx+cosx)(-sinx)]}/cos^2x
    = (cos^2x-sinxcosx)-(-sin^2x-sinxcosx)/cos^2x
    =(cos^2x-sinxcosx+sin^2x+sinxcosx/cos^2x
    =cos^2x+sin^2x/cos^2x
    = (cos^2x/cosx)+(sin^2x/cosx)
    = am i even close. thanks for the help
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Derivative of Trig Function using quotient rule Help

    Quote Originally Posted by psilver1 View Post
    y= sinx+cosx/cosx
    y'= f'g-fg'/g^2
    = {[(cosx-sinx)(cosx)]-[(sinx+cosx)(-sinx)]}/cos^2x
    = (cos^2x-sinxcosx)-(-sin^2x-sinxcosx)/cos^2x
    =(cos^2x-sinxcosx+sin^2x+sinxcosx/cos^2x
    =cos^2x+sin^2x/cos^2x
    This is correct. To simplify the expression use the identity \sin^2(x)+\cos^2(x)=1.

    Note that
    y = \frac{\sin(x)+\cos(x)}{\cos(x)} = \frac{\sin(x)}{\cos(x)}+\frac{\cos(x)}{\cos(x)} = \tan(x)+1
    and \frac{d}{dx}y = \frac{d}{dx}(\tan x+1) = \frac{1}{\cos^2(x)}
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