Use implicit differentiation and the power rule for integer powers to prove the power rule for rational powers.

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" $$\begin{gathered} q{y}^{q-1}\frac{dy}{dx}=p{x}^{p-1} \\ \Rightarrow \frac{dy}{dx}=\frac{p{x}^{p-1}}{q{y}^{q-1}} \\ \Rightarrow \frac{dy}{dx}=\frac{p}{q}\times \frac{x^{p-1}}{y^{q-1}} \\ \Rightarrow \frac{dy}{dx}=\frac{p}{q}{x}^{p-1}{y}^{-(q-1)} \end{gathered}$$

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

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

This is the required value.

Hence power rule for rational powers proved.