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Thread: 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 $\displaystyle (\sec{x})'=\sec{x}\tan{x}$, now to find the second derivative just use carefully the product rule.
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  3. #3
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    $\displaystyle 1.)(\sec(x))''$
    $\displaystyle = (\sec(x)\tan(x))'$
    $\displaystyle = \sec(x)'\tan(x) + '\sec(x)\tan(x)$
    $\displaystyle = \sec(x)\sec^2(x) + \sec(x)\tan(x)\tan(x)$
    $\displaystyle = \sec^3{x} + \sec{x}\tan^2{x}$

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