Using mathematical induction prove that d dx (x^n)=nx^ n-1 for al… - Vidyadip

Question

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

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Answer

To prove: $\text{P(n)}:\frac{\text{d}}{\text{dx}}\text{(x}^\text{n})=\text{nx}^{\text{n}-1}$ for all positive integers n
For n = 1,
$\text{P}(1):\frac{\text{d}}{\text{dx}}\text{(x)}=1=1.\text{x}^{1-1}$
$\therefore\ $P(n) is true for n = 1
Let P(k) is true for some positive integer k.
That is, $\text{P(k)}:\frac{\text{d}}{\text{dx}}\text{(x}^\text{k})=\text{kx}^{\text{k}-1}$
It has to be proved that P(k + 1) is also true.
Consider $\frac{\text{d}}{\text{dx}}(\text{x}^{\text{k}+1})=\frac{\text{d}}{\text{dx}}(\text{x}.\text{x}^\text{k})$
$=\text{x}^\text{k}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{d}}{\text{dx}}(\text{x}^\text{k})$ $[\text{By applying product rule}]$
$=\text{x}^\text{k}.1+\text{x}.\text{k}.\text{x}^{\text{k}-1}$
$=\text{x}^\text{k}+\text{kx}^\text{k}$
$=(\text{k}+1).\text{x}^\text{k}$
$=(\text{k}+1).\text{x}^{\text{(k}+1)-1}$
Thus, P(k + 1) is true whenever P(k) is true.
Therefore, by the principle of mathematical lnduction, the statement P(n) is true for every positive integer n.
Hence, proved.

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