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Thread: quick differentiation question about sin^2(x)

  1. #1
    Red
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    quick differentiation question about sin^2(x)

    Ignorant question here, but I was looking at this example i have, needing to take the derivative:

    f'(x) = sin^2(x) + cos(x)
    = 2sin(x)*d/dx(sin x) - sin(x)
    = 2sin(x)*cos(x) - sin(x)

    and so on... I'm wanting to know why the 2sin(x) isn't where the differentiation stops for sin^2(x)? there's the 2sin(x), and then the d/dx(sinx) added in as well. The cos(x) on the right side doesn't have any extra steps for it. Hope this makes sense, and i realize this is a silly question but I just wanted the clarification.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    You can consider this function as follows:

    $\displaystyle f(x) = (sin(x))^2 + cos(x)$

    Then,

    $\displaystyle f'(x) = 2(sin(x)).cos(x) - sin(x) = 2sin(x)cos(x) - sin(x)$

    Just as the equation:

    $\displaystyle y = (x^2+2)^2$

    becomes:

    $\displaystyle y' = 2(x^2 + 2). (2x) = 4x(x^2 + 2)$
    Last edited by Unknown008; Jun 24th 2010 at 08:13 AM. Reason: Forgot a math tag
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    $\displaystyle f(x)=sin^2(x)$

    $\displaystyle f'(x)=2sin(x)\cdot (sinx)'=2sin(x)\cdot cos(x)=sin(2x)$
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  4. #4
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    Quote Originally Posted by Red View Post
    I'm wanting to know why the 2sin(x) isn't where the differentiation stops for sin^2(x)?
    Because it's not true that $\displaystyle \left\{(\sin{x})^n\right\}' = n(\sin{x})^{n-1}$. In general, $\displaystyle \left\{[f(x)]^n\right\}' \ne n(f(x))^{n-1}$ unless $\displaystyle f(x) = x.$ The derivative of f at x is defined as the limit $\displaystyle \displaystyle f'(x)=\lim_{\delta{x}\to 0}\dfrac{f(x+\delta{x})-f(x)}{\delta{x}}}$, but unfortunately, many students (at school, anyway) understand it as $\displaystyle \left(x^n\right)'=nx^{n-1}$. The power rule is itself as a result of considering the function $\displaystyle f(x) = x^n$ at $\displaystyle x$, and using the definition of the derivative to calculate $\displaystyle \displaystyle \lim_{\delta{x}\to 0}\dfrac{f(x+\delta{x})^n-x^n}{\delta{x}}}$ as $\displaystyle f'(x)$. If I were to ever teach (school) calculus, it's one of those things I would write in giant letters somewhere above the board.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    You can write only the words: "By definition"
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