Answer
Proof by Induction
Work Step by Step
The Base Case: The $n$ is a positive integer. So the smallest value it can assume is 1 which we take to be our base case.
For $n=1$, f(x) = $x^1$
LHS = $f^{'}(x)$ = $(x)^{'}$ = 1 as is known that the derivative of the identity function is the constant unit function.
And RHS = $1 \times x^0$ = 1
Therefore, LHS = RHS
Inductive Hypothesis: Let's say for a given n = k,
$(x^k)^{'}$ = $k \times x^{k-1}$ ........................ (1)
So for k+1, LHS = $(x^{k+1})^{'}$ = $(x\times x^k)^{'}$
By product rule, LHS = $x\times (x^k)^{'} + x^{'} \times x^k $
By (1), LHS = $x\times (k \times x^{k-1} ) + 1 \times x^k $
or $LHS = k \times x^k + x^k = (k+1) \times x^k$ = RHS
Thus, by the domino effect of the principle of induction, the assertion is proved.
