Browse Questions

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

Toolbox:
• By using the property of mathematical induction,we prove $P(1)$- to be true $P(k)$ and $P(k+1)$ to be true when $n=1,k,k+1$.
Step 1:
Let $P(n)=\large\frac{d}{dx}$$(x^{\large n})=nx^{\large n-1}------(1) To verify the statement for n=1. Put n=1 in (1) we get P(1)=\large\frac{d}{dx}$$(x^1)=(1)x^{1-1}$
$\qquad\qquad\quad\;\;\;=(1)x^0$
$\qquad\qquad\quad\;\;\;=(1)(1)$
$\qquad\qquad\quad\;\;\;=1$
Which is true as $\large\frac{d}{dx}$$(x)=1 Suppose P(x) is true for n=m. P(m)=\large\frac{d}{dx}$$(x^{\large m})=mx^{\large m-1}$
Step 2:
We establish the truth of $P(m+1)$ we prove