Question

$x^{\frac{2}{3}}\left(\text{Differentiate with respect to}x,\text{using first principle}\right)$

Answer 1

Find derivative by first principle method

Let $f\left(x\right)=x^{\frac{2}{3}}$

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(x+h{)}^{\frac{2}{3}}-x^{\frac{2}{3}}}{h}$

$\left[f^{\prime }\right(x)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}]$

$={lim}_{h\to 0}\frac{x^{\frac{2}{3}}{(1+\frac{h}{x})}^{\frac{2}{3}}-x^{\frac{2}{3}}}{h}$

$={lim}_{h\to 0}\frac{x^{\frac{2}{3}}[{(1+\frac{h}{x})}^{\frac{2}{3}}-1]}{h}$

$={lim}_{h\to 0}\frac{x^{\frac{2}{3}}[1+\frac{2}{3}\frac{h}{x}-\frac{1}{9}\frac{h^2}{x^2}+\dots -1]}{h}$

$\left[\begin{array}{c}(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\\ \dots +\frac{n(n-1)\dots (n-r+1)}{r!}{x}^r\end{array}\right]$

$={lim}_{h\to }{x}^{\frac{2}{3}}[\frac{2}{3x}-\frac{h}{9x^2}+\dots ]$

Now, apply the limit, we get

$f^{\prime }\left(x\right)=\frac{2}{3}{x}^{-\frac{1}{3}}$

Hence the required answer is $\frac{2}{3}{x}^{-\frac{1}{3}}$