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Math Help - Trig differentiation.

  1. #1
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    Trig differentiation.

    Hi,
    I'm having a lot of difficulty differentiating the following trig problems:

    Question : Find \frac{dy}{dx} for the following functions(Do not simplify the answer).

    a) 4 cos^2(\pi x)

    b) tan^2 (cos x)

    c) sin x +sin y=1

    I'm not even sure about how to start on these problems, and would greatly appreciate any helpful tips.


    Thanks
    Last edited by mr fantastic; June 5th 2010 at 04:43 PM. Reason: Edited post title.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by spoc21 View Post
    Hi,
    I'm having a lot of difficulty differentiating the following trig problems:

    Question : Find \frac{dy}{dx} for the following functions(Do not simplify the answer).

    a) 4 cos^2(\pi x)

    b) tan^2 (cos x)

    c) sin x +sin y=1

    I'm not even sure about how to start on these problems, and would greatly appreciate any helpful tips.


    Thanks
    Well,

    1: y = 4 cos^2(\pi x)

     = 4 cos( \pi x ) cos( \pi x )

    Then,

     \frac{dy}{dx} = 4[- \pi sin( \pi x) cos( \pi x) - \pi sin( \pi x) cos ( \pi x ) ]


    3: sin x +sin y=1

     cosx + cosy y^{ \prime } = 0

     y^{ \prime} = - \frac{cosx}{cosy}

    Compute 2 in the same way as 1.
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  3. #3
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    Quote Originally Posted by spoc21 View Post
    Hi,
    I'm having a lot of difficulty differentiating the following trig problems:

    Question : Find \frac{dy}{dx} for the following functions(Do not simplify the answer).

    a) 4 cos^2(\pi x)

    b) tan^2 (cos x)

    c) sin x +sin y=1

    I'm not even sure about how to start on these problems, and would greatly appreciate any helpful tips.


    Thanks
    a)

    y=4cos^2({\pi}x)

    u={\pi}x

    v=cosu

    w=v^2

    y=4w

    By the chain rule...

    \frac{dy}{dx}=\frac{dy}{dw}\ \frac{dw}{dv}\ \frac{dv}{du}\ \frac{du}{dx}


    b)

    similarly


    c)

    \frac{d}{dx}\ sinx+\frac{d}{dx}siny=\frac{d}{dx}1=0

    cosx+\frac{dy}{dx}\ \frac{d}{dy}siny=0

    \frac{dy}{dx}(cosy)=-cosx

    \frac{dy}{dx}=-\frac{cosx}{cosy}
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