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Math Help - finding the second derivitive

  1. #1
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    finding the second derivitive

    I had some problems trying to find the second derivative of some equations.
    1. sec(x). I used the product rule when i got to the first deriv. but my answer came as sec x^3 which is incorrect and it says: sec(x)^3+secx*tanx^2 as the right answer.
    2. e^x*cos x. Again I used the product rule and i got -e^x*cos x+e^x*cos x which is wrong and it says -2e^x*sin x is the right answer.
    3.(sqrt x) I had a problem with the distribution and exponents. Anyways, I got x^1/2*(1/x)+(1/2 x^-1/2)*ln x. the right answer is -1/4*x^-3/2 *ln t.
    Please help.
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  2. #2
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    Quote Originally Posted by driver327 View Post
    1. sec(x). I used the product rule when
    Product rule?

    This is a basic derivative, which is (\sec{x})'=\sec{x}\tan{x}, now to find the second derivative just use carefully the product rule.
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  3. #3
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    1.)(\sec(x))''
    = (\sec(x)\tan(x))'
    = \sec(x)'\tan(x) + '\sec(x)\tan(x)
    = \sec(x)\sec^2(x) + \sec(x)\tan(x)\tan(x)
    = \sec^3{x} + \sec{x}\tan^2{x}

    2.)(e^{x}\cos(x))''
    = (-e^{x}sin{x} + e^{x}cos{x})'
    = -e^{x}\sin{x} -e^{x}\cos + e^x\cos{x} - e^{x}\sin{x}
    cancel the cosine
    = -2e^{x}\sin{x}
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  4. #4
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    3.) (\sqrt{x})''
    =(\frac{1}{2\sqrt{x}})'
    =\frac{1}{2}(x^{-\frac{1}{2}})
    =\frac{1}{2}(-\frac{1}{2}(x^{-\frac{1}{2}-1}))
    =-\frac{1}{4}(x^{-\frac{3}{2}})
    =-\frac{1}{4x^{\frac{3}{2}}}
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