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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^(r1) by logarithmic differentiation
anonymous
 one year ago
if r is any real number and y=x^r is a differentiable function Show that dy/dx=rx^(r1) by logarithmic differentiation

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx }=rx ^{r1}\]

phi
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a website, or a textbook with is proof will be helpful thank ..or a video ....anyone??

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Suppose 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 \[\large \frac{d}{dx} (x^r) = x^r [(r)x^{1}]\] \[\large \frac{d}{dx} (x^r) = x^rx^{1} [(r)]\] \[\large \frac{d}{dx} (x^r) = x^{r1} [(r)]\] 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 \[\large \frac{dy}{dx} = rx^{r1}\] only what I've done is rearranged differently with the x being first and then r.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2._. I think....... proofs kill me >_< but I tried :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that by by logarithmic differentiation? and thanks

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2that should be through logarithmic differentiation but I gotta be honest... towards the end felt like power rule, but that's how I got through : /

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thanks for you time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see u like me @Mehek14 hehe
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