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Thread: Derivative

  1. #1
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    Derivative

    Hello,

    can someone please show me how to get the first and second derivative for

    $\displaystyle 2cosx+cos^2x$
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  2. #2
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    Quote Originally Posted by l flipboi l View Post
    Hello,

    can someone please show me how to get the first and second derivative for

    $\displaystyle 2cosx+cos^2x$
    hi flipboi,

    The derivative of $\displaystyle Cosx$ is $\displaystyle -Sinx$,
    so the derivative of $\displaystyle 2Cosx$ is the derivative of $\displaystyle (Cosx+Cosx)$ which is $\displaystyle (-Sinx-Sinx)=-2Sinx$.

    If you use the trigonometric identity $\displaystyle Cos^2x=\frac{1}{2}(1+Cos2x)$ you can more easily find the derivative of $\displaystyle Cos^2x.$

    $\displaystyle \frac{d}{dx}\frac{1}{2}(1+Cos2x)=\frac{1}{2}\frac{ d}{dx}Cos2x$

    since the derivative of a constant is zero.

    Now you have to use the Chain Rule to complete.

    $\displaystyle u=2x,\ \frac{du}{dx}=2$

    $\displaystyle \frac{d}{dx}Cos2x=\frac{d}{dx}Cosu=\frac{du}{dx}\f rac{d}{du}Cosu=2(-Sinu)=-2Sin2x$

    Therefore the derivative of $\displaystyle Cos^2x$ is $\displaystyle -Sin2x$

    Combine the two for your final answer.
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    hi flipboi,

    The derivative of $\displaystyle Cosx$ is $\displaystyle -Sinx$,
    so the derivative of $\displaystyle 2Cosx$ is the derivative of $\displaystyle (Cosx+Cosx)$ which is $\displaystyle (-Sinx-Sinx)=-2Sinx$.

    If you use the trigonometric identity $\displaystyle Cos^2x=\frac{1}{2}(1+Cos2x)$ you can more easily find the derivative of $\displaystyle Cos^2x.$

    $\displaystyle \frac{d}{dx}\frac{1}{2}(1+Cos2x)=\frac{1}{2}\frac{ d}{dx}Cos2x$

    since the derivative of a constant is zero.

    Now you have to use the Chain Rule to complete.

    $\displaystyle u=2x,\ \frac{du}{dx}=2$

    $\displaystyle \frac{d}{dx}Cos2x=\frac{d}{dx}Cosu=\frac{du}{dx}\f rac{d}{du}Cosu=2(-Sinu)=-2Sin2x$

    Therefore the derivative of $\displaystyle Cos^2x$ is $\displaystyle -Sin2x$

    Combine the two for your final answer.
    For the first derivative I did this:

    -2sin(x) - 2cos(x)sin(x)

    would this be correct?

    for the second, i'm having a bit trouble.
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  4. #4
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    Yes,

    you expressed the double angle in terms of the single angle.
    To obtain the second derivative, just treat the CosxSinx as a product
    and differentiate using the product rule.

    $\displaystyle \frac{d}{dx}(-2Sinx-2CosxSinx)=-2\frac{d}{dx}(Sinx+CosxSinx)$

    $\displaystyle =-2\frac{d}{dx}Sinx-2\frac{d}{dx}uv=-2\frac{d}{dx}Sinx-2(v\frac{du}{dx}+u\frac{dv}{dx})$

    $\displaystyle =-2\frac{d}{dx}Sinx-2(Sinx\frac{d}{dx}Cosx+Cosx\frac{d}{dx}Sinx)$
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    Yes,

    you expressed the double angle in terms of the single angle.
    To obtain the second derivative, just treat the CosxSinx as a product
    and differentiate using the product rule.

    $\displaystyle \frac{d}{dx}(-2Sinx-2CosxSinx)=-2\frac{d}{dx}(Sinx+CosxSinx)$

    $\displaystyle =-2\frac{d}{dx}Sinx-2\frac{d}{dx}uv=-2\frac{d}{dx}Sinx-2(v\frac{du}{dx}+u\frac{dv}{dx})$

    $\displaystyle =-2\frac{d}{dx}Sinx-2(Sinx\frac{d}{dx}Cosx+Cosx\frac{d}{dx}Sinx)$
    so the 2nd derivative would be?

    $\displaystyle -2cosx - 2cos^2x+2sin^2x
    $

    is there anyway to simplify this?
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  6. #6
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    Yes flipboi,

    You could simplify a little, using $\displaystyle Cos^2x-Sin^2x=Cos(2x)$

    Then, $\displaystyle -2Cosx-2Cos^2x+2Sin^2x=-2Cosx-2(Cos^2x-Sin^2x)$

    $\displaystyle =-2Cosx-2Cos2x$
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  7. #7
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    Thats what I was missing! what is it called when you convert cos^2x - sin^2x = cos2x?
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  8. #8
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    That's one of the list of trigonometric identities.

    It comes from $\displaystyle Cos(A+B)=CosACosB-SinASinB$

    If A and B are both x, we get $\displaystyle Cos(x+x)=Cos2x=CosxCosx-SinxSinx=Cos^2x-Sin^2x$
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  9. #9
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    You can also write

    $\displaystyle Cos2x=2Cos^2x-1\ \Rightarrow\ -2Cos2x=-4Cos^2x+2$

    giving $\displaystyle -2Cosx-2Cos2x=-2Cosx-4Cos^2x+2=-2Cosx(1+2Cosx)+2$
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