## 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

1. anonymous

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

2. 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

3. anonymous

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

4. anonymous

@UsukiDoll

5. 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.

6. UsukiDoll

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

7. anonymous

is that by by logarithmic differentiation? and thanks

8. 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 : /

9. anonymous

ok thanks for you time

10. anonymous

I see u like me @Mehek14 hehe