Q 94 - Calculus

Question

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

Step-by-Step Solution

Answer

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

Hence power rule for rational powers proved.

Step 1. Given Information

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.

Step 2. Prove the power rule for rational powers

Consider the following function :- 

$y=x^{p/q}$

We can write it as :- 

$y^q=x^p$

Take differentiation on both sides, the we have :-

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

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

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

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

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

This is the required value.

Hence power rule for rational powers proved.