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Math Help - derivative

  1. #1
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    derivative

    y = sec(logx)

    y' = sec(logx)tan(logx)*1/secx

    correct?

    thank u
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  2. #2
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    almost, if in doubt use the chain rule..

    y = \sec ( \log x) Let u = \log x

    y = \sec u \ \ \rightarrow \ \  \frac{dy}{du}= \sec u \tan u

    \frac{du}{dx} = \frac{1}{x}

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


    Bobak
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  3. #3
    Member Greengoblin's Avatar
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    You have f(x)=\sec(x), g(x)=\ln(x), and\ f\circ g(x) = \sec(\ln(x))

    So you need to use the chain rule:
     <br /> <br />
\frac{dy}{dx}=\frac{df}{dg} \frac{dg}{dx} = \sec(g(x))\tan(g(x)) \frac{1}{x} = \frac{\sec(ln(x))\tan(ln(x))}{x}<br />
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  4. #4
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    Quote Originally Posted by Greengoblin View Post
    You have f(x)=\sec(x), g(x)=\ln(x), and\ f\circ g(x) = \sec(\ln(x))

    So you need to use the chain rule:
     <br /> <br />
\frac{dy}{dx}=\frac{df}{dg} \frac{dg}{dx} = \sec(g(x))\tan(g(x)) \frac{1}{x} = \frac{\sec(ln(x))\tan(ln(x))}{x}<br />

    thanks for that.
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