Answer
$\frac{d}{dx} x^{r} = rx^{r-1}$ was proven using implicit differentiation.
Work Step by Step
Let $y = x^{r} = x^{\frac{m}{n}}$. Therefore, $y^n = x^m$.
Taking the derivative of both sides, we get:
$ny^{n-1}\frac{dy}{dx} = mx^{m-1}$
$\frac{dy}{dx} = \frac{mx^{m-1}}{ny^{n-1}}$
$\frac{dy}{dx} = \frac{mx^{m-1}}{n(x^{\frac{m}{n}})^{n-1}}$
$\frac{dy}{dx} = \frac{mx^{m-1}}{n(x^{m-m/n})}$
$\frac{dy}{dx} = \frac{m}{n}x^{(m-1)-(m-m/n)}$
$\frac{dy}{dx} = \frac{m}{n}x^{\frac{m}{n}-1}$
$\frac{dy}{dx} = rx^{r-1}$
Thus, $\frac{d}{dx} x^{r} = rx^{r-1}$ was proven using implicit differentiation.
