Question

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

Answer 1

Check the given expression for $n=1$

L.H.S. $=\frac{d}{dx}\left(x^1\right)=\frac{dx}{dx}=1$

R.H.S. $=1.x^{1-1}=1$

Hence, it is true for $n=1$

Let the expression be true for $n=k$

$\frac{d}{dx}\left(x^k\right)=k{x}^{k-1}$

Now, check for $n=k+1$

L.H.S. $=\frac{d}{dx}\left(x^{k+1}\right)=\frac{d}{dx}(x^k.x)=x\frac{d}{dx}\left(x^k\right)+x^k\frac{d}{dx}\left(x\right)=x\left(k{x}^{k-1}\right)+x^k\left(1\right)=k{x}^k+x^k=(k+1)x^k=(k+1)x^{(k+1)-1}$

Hence, the given expression is true for $n=k+1$ too.

By principle of mathematical induction, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$