anonymous
  • anonymous
if r is any real number and y=x^r is a differentiable function Show that dy/dx=rx^(r-1) by logarithmic differentiation
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{ dy }{ dx }=rx ^{r-1}\]
phi
  • phi
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
anonymous
  • anonymous
a website, or a textbook with is proof will be helpful thank ..or a video ....anyone??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@UsukiDoll
UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
._. I think....... proofs kill me >_< but I tried :/
anonymous
  • anonymous
is that by by logarithmic differentiation? and thanks
UsukiDoll
  • UsukiDoll
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 : /
anonymous
  • anonymous
ok thanks for you time
anonymous
  • anonymous
I see u like me @Mehek14 hehe

Looking for something else?

Not the answer you are looking for? Search for more explanations.