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anonymous

  • one year ago

if r is any real number and y=x^r is a differentiable function Show that dy/dx=rx^(r-1) by logarithmic differentiation

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  1. anonymous
    • one year ago
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    \[\frac{ dy }{ dx }=rx ^{r-1}\]

  2. phi
    • one year ago
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    I think they want you to first take the log of both sides \[ \ln y = \ln(x^r) \] and use implicit differentiation to find dy/dx

  3. anonymous
    • one year ago
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    a website, or a textbook with is proof will be helpful thank ..or a video ....anyone??

  4. anonymous
    • one year ago
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    @UsukiDoll

  5. UsukiDoll
    • one year ago
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    Suppose we have \[\large f(x) = x^r \] with r being the real number and since we need to use logarithmic differentiation . maybe if we let \[\large x^r = e^{r \cdot \ln(x)}\] then take the derivative with the exponential and logarithmic differentiation \[\large \frac{d}{dx} (x^r) = e^{r \cdot \ln(x)} (r \cdot \ln(x))'\] next substituting back \[\large x^r = e^{r \cdot \ln(x)}\] and using product rule we have \[\large \frac{d}{dx} (x^r) = x^r (r \cdot \ln(x))'\] \[\large \frac{d}{dx} (x^r) = x^r [(r)(\frac{1}{x})+ ln(x)(0)]\] \[\large \frac{d}{dx} (x^r) = x^r [(r)(\frac{1}{x})]\] rewriting 1/x and rearranging \[\large \frac{d}{dx} (x^r) = x^r [(r)x^{-1}]\] \[\large \frac{d}{dx} (x^r) = x^rx^{-1} [(r)]\] \[\large \frac{d}{dx} (x^r) = x^{r-1} [(r)]\] Therefore we have proven that if r is a real number and \[\large y=x^r \] is a differential function, then we can use logarithmic differentiation and obtain \[\large \frac{dy}{dx} = rx^{r-1}\] only what I've done is rearranged differently with the x being first and then r.

  6. UsukiDoll
    • one year ago
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    ._. I think....... proofs kill me >_< but I tried :/

  7. anonymous
    • one year ago
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    is that by by logarithmic differentiation? and thanks

  8. UsukiDoll
    • one year ago
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    that should be through logarithmic differentiation but I gotta be honest... towards the end felt like power rule, but that's how I got through : /

  9. anonymous
    • one year ago
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    ok thanks for you time

  10. anonymous
    • one year ago
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    I see u like me @Mehek14 hehe

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