91Ó°ÊÓ

We have to prove the power rule for rational powers.

That is we have to prove that :-

$\frac{d}{dx}\left(x^{p/q}\right)=\frac{p}{q}{x}^{p/q-1}$

We have to use implicit differentiation and the power rule for integer powers to prove this thing.

Consider the following function :-

$y=x^{p/q}$

We can write it as :-

localid="1648654837463" $y^q=x^p$

Take differentiation on both sides, the we have :-

localid="1648654859796" $\frac{d}{dx}\left(y^q\right)=\frac{d}{dx}\left(x^p\right)$

Then by applying power rule for integer powers and chain rule, we have :-

localid="1648654884669" $$\begin{gathered} q{y}^{q-1}\frac{dy}{dx}=p{x}^{p-1} \\ ⇒\frac{dy}{dx}=\frac{p{x}^{p-1}}{q{y}^{q-1}} \\ ⇒\frac{dy}{dx}=\frac{p}{q}×\frac{x^{p-1}}{y^{q-1}} \\ ⇒\frac{dy}{dx}=\frac{p}{q}{x}^{p-1}{y}^{-(q-1)} \end{gathered}$$

Put the value of $y=x^{p/q}$, then we have :-

localid="1648654975215" $$\begin{gathered} ⇒\frac{d}{dx}\left(x^{p/q}\right)=\frac{p}{q}{x}^{p-1}{\left(x^{p/q}\right)}^{-(q-1)}⇒\frac{d}{dx}\left(x^{p/q}\right)=\frac{p}{q}{x}^{p-1}{x}^{-\left(p-p/q\right)} \\ ⇒\frac{d}{dx}\left(x^{p/q}\right)=\frac{p}{q}{x}^{p-1}{x}^{p/q-p} \\ ⇒\frac{d}{dx}\left(x^{p/q}\right)=\frac{p}{q}{x}^{p-1+p/q-p} \\ ⇒⇒\frac{d}{dx}\left(x^{p/q}\right)=\frac{p}{q}{x}^{p/q-1} \end{gathered}$$

This is the required value.

Hence power rule for rational powers proved.