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Thread: Trigonometric Derivatives

  1. #1
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    Trigonometric Derivatives

    I have a problem with understanding derivatives.
    How do I find the derivative and also simplify it?

    $\displaystyle x cos x$

    $\displaystyle sec^2 3x$

    An explanation would really help for me to understand better.
    Thank you!

    --------
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  2. #2
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    Quote Originally Posted by meiyukichan View Post
    I have a problem with understanding derivatives.
    How do I find the derivative and also simplify it?

    $\displaystyle x cos x$

    ...
    You are dealing with a product of 2 functions:

    $\displaystyle f(x)=x \cdot \cos(x)$ ....... Use product rule to calculate the drivation:

    If $\displaystyle f(x)=g(x) \cdot h(x)$ ....... then ....... $\displaystyle f'(x)=h(x) \cdot g'(x) + g(x) \cdot h'(x)$

    Therefore:

    $\displaystyle f(x)=x \cdot \cos(x)~\implies~f'(x)=\cos(x) \cdot 1 + x\cdot (-\sin(x))~\implies~f'(x)=\cos(x)-x\sin(x)$
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by meiyukichan View Post
    I have a problem with understanding derivatives.
    How do I find the derivative and also simplify it?

    $\displaystyle x cos x$

    $\displaystyle sec^2 3x$

    An explanation would really help for me to understand better.
    Thank you!

    --------
    I also need help on these derivative problems: CLICK
    You need four facts, namely:

    $\displaystyle \frac{d}{dx}\bigg[f(g(x))\bigg]=f'(g(x))\cdot{g'(x)}$

    $\displaystyle \frac{d}{dx}\bigg[f(x)\cdot{g(x)}\bigg]=f'(x)g(x)+g'(x)f(x)$

    $\displaystyle \frac{d}{dx}\bigg[\cos(x)\bigg]=-\sin(x)$

    $\displaystyle \frac{d}{dx}\bigg[\sec(x)\bigg]=\sec(x)\tan(x)$

    Now use those in combination to attaing the desired derivatives.
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    You need four facts, namely:

    $\displaystyle \frac{d}{dx}\bigg[f(g(x))\bigg]=f'(g(x))\cdot{g'(x)}$

    $\displaystyle \frac{d}{dx}\bigg[f(x)\cdot{g(x)}\bigg]=f'(x)g(x)+g'(x)f(x)$

    $\displaystyle \frac{d}{dx}\bigg[\cos(x)\bigg]=-\sin(x)$

    $\displaystyle \frac{d}{dx}\bigg[\sec(x)\bigg]=\sec(x)\tan(x)$

    Now use those in combination to attaing the desired derivatives.
    I don't understand what you mean by that.
    Can you show me how in an example similar to $\displaystyle sec^2 3x$
    or you can just use that as an example because it'll be easy for me to understand for similar problems.
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  5. #5
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    Quote Originally Posted by meiyukichan View Post
    I don't understand what you mean by that.
    Can you show me how in an example similar to $\displaystyle sec^2 3x$
    or you can just use that as an example because it'll be easy for me to understand for similar problems.
    This is a great website for finding derivatives: http://www.mathhelpforum.com/math-he...step-step.html

    The integration version is just as good for checking answers but it doesn't give you step-by-step solutions.
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  6. #6
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    Quote Originally Posted by r_maths View Post
    This is a great website for finding derivatives: http://www.mathhelpforum.com/math-he...step-step.html

    The integration version is just as good for checking answers but it doesn't give you step-by-step solutions.
    Okay. Thank you for the advice/tip/help!
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