Q 92 - Calculus

Question

Use implicit differentiation and the power rule for integer powers (not the general power rule) to prove that 

Step-by-Step Solution

Answer

$$\begin{gathered} y=x^{2/3} \\ \Rightarrow y^3=x^2 \\ \Rightarrow \frac{d}{dx}{y}^3=\frac{d}{dx}{x}^2 \\ \Rightarrow 3y^2\frac{dy}{dx}=2x \\ \Rightarrow \frac{dy}{dx}=\frac{2x}{3y^2} \\ \Rightarrow \frac{dy}{dx}=\frac{2}{3}x{y}^{-2} \\ \Rightarrow \Rightarrow \frac{d}{dx}\left(x^{2/3}\right)=\frac{2}{3}x{\left(x^{2/3}\right)}^{-2} \\ \Rightarrow \frac{d}{dx}\left(x^{2/3}\right)=\frac{2}{3}x\left(x^{-4/3}\right) \\ \Rightarrow \frac{d}{dx}\left(x^{2/3}\right)=\frac{2}{3}{x}^{1-4/3} \\ \Rightarrow \frac{d}{dx}\left(x^{2/3}\right)=\frac{2}{3}{x}^{-1/3} \end{gathered}$$

Hence proved.

Step 1. Given Information

We have given that :-

$\frac{d}{dx}\left(x^{2/3}\right)=\frac{2}{3}{x}^{-1/3}$
We have to prove this derivative by using implicit differentiation and power rule for integer powers.

Step 2. Prove that d d x ( x 2 / 3 ) = 2 3 x - 1 / 3

We have to find the derivative of  $x^{2/3}$.

Let :-

$y=x^{2/3}$.

We can write it as :-

$y^3=x^2$.

Then by using implicit differentiation :-

$\frac{d}{dx}{y}^3=\frac{d}{dx}{x}^2$

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

$$\begin{gathered} 3y^2\frac{dy}{dx}=2x \\ \Rightarrow \frac{dy}{dx}=\frac{2x}{3y^2} \\ \Rightarrow \frac{dy}{dx}=\frac{2}{3}x{y}^{-2} \end{gathered}$$

Put the value $y=x^{2/3}$, then we have :-

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(x^{2/3}\right)=\frac{2}{3}x{\left(x^{2/3}\right)}^{-2} \\ \Rightarrow \frac{d}{dx}\left(x^{2/3}\right)=\frac{2}{3}x\left(x^{-4/3}\right) \\ \Rightarrow \frac{d}{dx}\left(x^{2/3}\right)=\frac{2}{3}{x}^{1-4/3} \\ \Rightarrow \frac{d}{dx}\left(x^{2/3}\right)=\frac{2}{3}{x}^{-1/3} \end{gathered}$$

This is the required value.

Hence proved.