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Math Help - Differentiate help!

  1. #1
    Rui
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    Differentiate help!

    a) f(x) = (√x + (1/ x^1/3))^2
    b) f(x) = √x ( 1+ sin^2 (2π/x))
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  2. #2
    Moo
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    Hello,

    a) f(x) = (√x + (1/ x^1/3))^2
    \left(\sqrt{x}+\left(\frac{1}{x^{1/3}}\right)\right)^2=\left(\sqrt{x}+x^{-1/3}\right)^2

    You have something like (u(x))^2. Its derivative is 2u'(x)u(x)

    Here, u'(x) is the derivative of a sum.

    u'(x)=\frac{1}{2\sqrt{x}}+(-1/3)x^{-4/3}

    Hence f'(x)=\dots
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  3. #3
    Moo
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    Quote Originally Posted by Rui View Post
    b) f(x) = √x ( 1+ sin^2 (2π/x))
    Use the product and the chain rule
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  4. #4
    Member Danshader's Avatar
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    hello there!,

    well i hope you know your basic rules of differentiation:
    \frac{d}{dx}x^n = nx^{(n-1)}
    \frac{d}{dx}x^2 = 2x^{(2-1)} = 2x
    \frac{d}{dx}(x+x^2)^n = n(x+x^2)^{(n-1)}\frac{d}{dx}(x+x^2)

    i will solve question a) for you and you will try to answer question b)

    <br />
\frac{d}{dx}f(x) =\frac{d}{dx} (  x^\frac{1}{2}+ x^{- \frac{1}{3}})^2<br />
    <br />
\frac{d}{dx}f(x) =2(  x^\frac{1}{2}+ x^{- \frac{1}{3}})^{(2-1)}\frac{d}{dx} (  x^\frac{1}{2}+ x^{- \frac{1}{3}})<br />
    <br />
\frac{d}{dx}f(x) =2(  x^\frac{1}{2}+ x^{- \frac{1}{3}}) \left ( \frac{1}{2}x^{(\frac{1}{2} - 1)} - \frac{1}{3}x^{(- \frac{1}{3} -1)} \right)<br />
    <br />
\frac{d}{dx}f(x) =2( x^\frac{1}{2}+ x^{- \frac{1}{3}}) \left ( \frac{1}{2}x^{-\frac{1}{2} } - \frac{1}{3}x^{- \frac{4}{3} } \right)<br />
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