Question

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

Answer 1

\[\frac{ dy }{ dx }=rx ^{r-1}\]

Answer 2

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

Answer 3

a website, or a textbook with is proof will be helpful thank ..or a video ....anyone??

Answer 4

@UsukiDoll

Answer 5

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.

Answer 6

._. I think....... proofs kill me >_<
but I tried :/

Answer 7

is that by by logarithmic differentiation? and thanks

Answer 8

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 : /

Answer 9

ok thanks for you time

Answer 10

I see u like me @Mehek14 hehe