Question

Using mathematical induction prove that d/dx.(x^n)=nx^n-1 for all positive integers n.

Answer 1

we have to prove :- $\bf{\frac{d}{dx}x^n=nx^{n-1}}$ for all integers n.

for n = 1
$\bf{LHS =\frac{d}{dx}(x)=1}$
RHS = 1.x^{1-1} = 1
So, LHS = RHS
∴ P(1) is true.
∴ P(n) is true for n = 1

Let P(k) be true for some positive integer k.
e.g., P(k) = $\bf{\frac{d}{dx}(x^k)=kx^{k-1}}$
Now, to prove that P(k + 1) is also true
RHS = $\bf{(k+1)x^{k+1-1}}$

LHS = $\bf{\frac{d}{dx}(x^{k+1})=\frac{d}{dx}(x\times x^k)}$

=$\bf{x^k\frac{d}{dx}(x)+x\frac{d}{dx}(x^k)}$

=$\bf{x^k.1+x.kx^{k-1}}$

=$\bf{x^{k+1-1}+kx^{k+1-1}}$

=$\bf{(k+1)x^{k+1-1}}$
∴ LHS = RHS
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.
$\textbf{\underline{hence proved}}$