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

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  • 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
$P(m+1)=\large\frac{d}{dx}$$(x^{\large m+1})=(m+1)x^{\large m}$
Since $x^{\large m+1}=x^{\large 1}.x^{\large m}$
$\large\frac{d}{dx}$$(x^{\large m+1})=\large\frac{d}{dx}$$(x.x^{\large m})$
$\qquad\qquad=x\large\frac{d}{dx}$$(x^{\large m})+x^{\large m}.\large\frac{d}{dx}$$(x)$
$\qquad\qquad=x.mx^{\large m-1}+x^m.1$
$\qquad\qquad=mx^{\large m}+x^{\large m}$
$\qquad\qquad=(m+1)x^{\large m}$
$\qquad\qquad=(m+1)x^{\large (m+1)-1}$
Step 3:
$P(m+1)$ is true if $P(m)$ is true.
By principle of induction $P(n)$ is true for all $n\in N$.
answered May 15, 2013 by sreemathi.v
 

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