Using mathematical induction prove that $\cfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$ for all positive integers it.

Let $P(n)$ be the given statement in the problem

$P(n):\cfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$ …(i)
For n $=$ 1,
Putting n $=$ 1 in (i), we get

$P(1):\cfrac{d}{{dx}}({x^1}) = (1){x^{1 - 1}} = (1){x^0} = (1)(1) = 1$
which is true as (x) $=$ 1

We suppose $P(n)$ is true for n $=$ m,
$P(m):\cfrac{d}{{dx}}({x^m}) = m{x^{m - 1}}$ …(ii)

To establish the truth of $P(m + 1)$, we have to prove

$P(m + 1):\cfrac{d}{{dx}}({x^{m + 1}}) = (m + 1){x^m}$ …(ii)

Now, ${x^{m + 1}} = {x^1} \cdot {x^m}$

$\Rightarrow$ $\cfrac{d}{{dx}}({x^{m + 1}}) = \cfrac{d}{{dx}}(x \cdot {x^m}) = x \cdot \cfrac{d}{{dx}}({x^m}) + {x^m}\cfrac{d}{{dx}}(x)$

$= x \cdot m{x^{m - 1}} + {x^m}$
$= m{x^m} + {x^m} = {x^m}(m + 1) = (m + 1){x^{(m + 1) - 1}}$

therefore, $P(n)$ is true for n$=$ m+ 1.

By using principle of induction, P(n) is true for all $n \in N.$