Question

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

Answer 1

To prove:P(n):\(\frac{d}{dx}\)(xⁿ)=nx^n-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.