Question

Using mathematical induction prove that $\frac{d}{dx}$(xn)=nxn-1 for all positive integers n

Answer 1

To prove:P(n):$\frac{d}{dx}$(xn)=nxn-1 for all positive integers n
For n=1
P(1)=$\frac{d}{dx}$(x)=1=1.x1-1
∴P(n) is true for n=1
Let P(k) be true for some positive integer k.
That is,P(k):$\frac{d}{dx}$(xk)=kxk-1
It has to be proved that P(k+1) is also true.
Consider $\frac{d}{dx}$(xk+1)=$\frac{d}{dx}$(x.xk)
=xk.$\frac{d}{dx}$(x)+x.$\frac{d}{dx}$(xk)
=xk.1+x.k.xk-1
=xk+kxk
=(k+1).xk
=(k+1).x(k+1)-1
Thus, P(k+1) is true whenever P(k) is true.
Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n.
Hence, it proved.