Question

find the derivative of x√x with respect to x using method of first principle​

Answer 1

The first principle of derivative states that the first derivative of the function $f(x)$ is given by,

$\displaystyle\longrightarrow f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

Here,

• $f(x)=x\sqrt x=x^{\frac{3}{2}}$

Then by first principle, its first derivative is,

$\displaystyle\longrightarrow f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

$\displaystyle\longrightarrow f'(x)=\lim_{h\to0}\dfrac{(x+h)^{\frac{3}{2}}-x^{\frac{3}{2}}}{h}$

$\displaystyle\longrightarrow f'(x)=\lim_{h\to0}\dfrac{\left[x\left(1+\frac{h}{x}\right)\right]^{\frac{3}{2}}-x^{\frac{3}{2}}}{h}$

$\displaystyle\longrightarrow f'(x)=\lim_{h\to0}\dfrac{x^{\frac{3}{2}}\left(1+\frac{h}{x}\right)^{\frac{3}{2}}-x^{\frac{3}{2}}}{h}$

$\displaystyle\longrightarrow f'(x)=x^{\frac{3}{2}}\lim_{h\to0}\dfrac{\left(1+\frac{h}{x}\right)^{\frac{3}{2}}-1}{h}\quad\quad\dots(1)$

Now, we have the following expansion.

$\longrightarrow(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}\,x^2+\dfrac{n(n-1)(n-2)}{3!}\,x^3+\,\dots$

Subtracting 1,

$\longrightarrow(1+x)^n-1=nx+\dfrac{n(n-1)}{2!}\,x^2+\dfrac{n(n-1)(n-2)}{3!}\,x^3+\,\dots$

Replacing $x$ by $\frac{h}{x}$ and taking $n=\frac{3}{2},$

$\small\text{ \longrightarrow\left(1+\dfrac{h}{x}\right)^{\frac{3}{2}}-1=\dfrac{3}{2}\left(\dfrac{h}{x}\right)+\dfrac{\frac{3}{2}\left(\frac{3}{2}-1\right)}{2!}\left(\dfrac{h}{x}\right)^2+\dfrac{\frac{3}{2}\left(\frac{3}{2}-1\right)\left(\frac{3}{2}-2\right)}{3!}\left(\dfrac{h}{x}\right)^3+\,\dots }$

Dividing by $h,$

$\small\text{ \longrightarrow\dfrac{\left(1+\frac{h}{x}\right)^{\frac{3}{2}}-1}{h}=\dfrac{3}{2x}+\dfrac{\frac{3}{2}\left(\frac{3}{2}-1\right)h}{2!\cdot x^2}+\dfrac{\frac{3}{2}\left(\frac{3}{2}-1\right)\left(\frac{3}{2}-2\right)h^2}{3!\cdot x^3}+\,\dots }$

Taking limit when $h\to0,$

$\displaystyle\small\text{ \longrightarrow\lim_{h\to0}\dfrac{\left(1+\frac{h}{x}\right)^{\frac{3}{2}}-1}{h}=\dfrac{3}{2x}+\dfrac{\frac{3}{2}\left(\frac{3}{2}-1\right)}{2!\cdot x^2}\lim_{h\to0}h+\dfrac{\frac{3}{2}\left(\frac{3}{2}-1\right)\left(\frac{3}{2}-2\right)}{3!\cdot x^3}\lim_{h\to0}h^2+\,\dots }$

$\displaystyle\small\text{ \longrightarrow\lim_{h\to0}\dfrac{\left(1+\frac{h}{x}\right)^{\frac{3}{2}}-1}{h}=\dfrac{3}{2x}+\dfrac{\frac{3}{2}\left(\frac{3}{2}-1\right)}{2!\cdot x^2}(0)+\dfrac{\frac{3}{2}\left(\frac{3}{2}-1\right)\left(\frac{3}{2}-2\right)}{3!\cdot x^3}(0)^2+\,\dots }$

$\displaystyle\longrightarrow\lim_{h\to0}\dfrac{\left(1+\frac{h}{x}\right)^{\frac{3}{2}}-1}{h}=\dfrac{3}{2x}$

Then (1) becomes,

$\displaystyle\longrightarrow f'(x)=x^{\frac{3}{2}}\cdot\dfrac{3}{2x}$

$\displaystyle\longrightarrow\underline{\underline{f'(x)=\sqrt x}}$