Question

d/dx multiplied by x to the power n=n.x to the power n-1


prove the equation.​

Answer 1

We're asked to prove,

$\longrightarrow \dfrac{d}{dx}(x^n)=nx^{n-1}\quad\forall n\in\mathbb{R}$

We're proving with the first principle of derivatives.

First principle of derivatives states that, the first derivative of a real valued function is given by,

$\displaystyle\large\longrightarrow \dfrac{d}{dx}[f(x)]=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

According to the question, let,

$\longrightarrow f(x)=x^n$

so that,

$\longrightarrow f(x+h)=(x+h)^n$

By first principle,

$\displaystyle\longrightarrow \dfrac{d}{dx}[f(x)]=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

$\displaystyle\longrightarrow \dfrac{d}{dx}(x^n)=\lim_{h\to0}\dfrac{(x+h)^n-x^n}{h}\quad\quad\dots(1)$

We are familiar with binomial theorem,

$\displaystyle\longrightarrow(a+b)^n=\sum_{r=0}^n\,^n\!C_r\ a^{n-r}\ b^r$

Taking $a=x$ and $b=h,$

$\displaystyle\longrightarrow(x+h)^n=\sum_{r=0}^n\,^n\!C_r\ x^{n-r}\ h^r$

We can take the first term out of the sum as,

$\displaystyle\longrightarrow(x+h)^n=\,(^nC_0\ x^{n-0}\ h^0)+\sum_{r=1}^n\,^n\!C_r\ x^{n-r}\ h^r$

$\displaystyle\longrightarrow(x+h)^n=x^n+\sum_{r=1}^n\,^n\!C_r\ x^{n-r+1-1}\ h^{r-1+1}$

$\displaystyle\longrightarrow(x+h)^n-x^n=\sum_{r=1}^n\,^n\!C_r\ x^{n-r+1}\cdot x^{-1}\cdot h^{r-1}\cdot h$

Taking $hx^{-1}$ out of the sum,

$\displaystyle\longrightarrow(x+h)^n-x^n=hx^{-1}\sum_{r=1}^n\,^n\!C_r\ x^{n-(r-1)}\,h^{r-1}$

$\displaystyle\longrightarrow\dfrac{(x+h)^n-x^n}{h}=x^{-1}\sum_{r=1}^n\,^n\!C_r\ x^{n-(r-1)}\,h^{r-1}$

Then (1) becomes,

$\displaystyle\longrightarrow\dfrac{d}{dx}(x^n)=\lim_{h\to0}x^{-1}\sum_{r=1}^n\,^n\!C_r\ x^{n-(r-1)}\,h^{r-1}$

Taking the first term out of the sum,

$\displaystyle\longrightarrow\dfrac{d}{dx}(x^n)=x^{-1}\cdot\lim_{h\to0}\left[\,^n\!C_1\ x^{n-(1-1)}\,h^{1-1}+\sum_{r=2}^n\,^n\!C_r\ x^{n-(r-1)}\,h^{r-1}\right]$

$\displaystyle\longrightarrow\dfrac{d}{dx}(x^n)=x^{-1}\cdot\left[\lim_{h\to0}nx^{n}+\lim_{h\to0}\sum_{r=2}^n\,^n\!C_r\ x^{n-(r-1)}\,h^{r-1}\right]$

$\displaystyle\longrightarrow\dfrac{d}{dx}(x^n)=x^{-1}\cdot\left[nx^{n}+0\right]$

$\displaystyle\longrightarrow\dfrac{d}{dx}(x^n)=x^{-1}\cdot nx^n$

$\displaystyle\longrightarrow\underline{\underline{\dfrac{d}{dx}(x^n)=nx^{n-1}}}$

Hence Proved!