Question

Prove that $\frac{d^n}{d{x}^n}(e^{2x}+e^{-2x}))=2^n[e^{2x}+(-1{)}^n(e^{-2x}\left)\right]$

Answer 1

We will prove this statement by mathematical induction.

We have $\frac{d}{dx}(e^{2x}+e^{-2x})=2[e^{2x}+(-1\left)e^{-2x}\right]$, so the given statement is true for $n=1$.

Again $\frac{d^2}{d{x}^2}(e^{2x}+e^{-2x})=\frac{d}{dx}\left\{2\right[e^{2x}+(-1)e^{-2x}\left]\right\}$$=2^2[e^{2x}+(-1{)}^2{e}^{-2x}]$. so the statement is also true for $n=2$ also.

Let us assume that the statement is true for $n=k$ i.e.

$\frac{d^k}{d{x}^k}$$(e^{2x}+e^{-2x})=$$2^k(e^{2x}+(-1{)}^n{e}^{-2x})$.......(1)

We will be proving that the statement is also true for $n=k+1$.

Now,

$\frac{d^{k+1}}{d{x}^{k+1}}$$(e^{2x}+e^{-2x})$

$=\frac{d}{dx}\frac{d^k}{d{x}^k}$$(e^{2x}+e^{-2x})$

$=\frac{d}{dx}\left\{2^k\right(e^{2x}+(-1{)}^k{e}^{-2x})\}$ [Using (1)]

$=2^{k+1}(e^{2x}+(-1{)}^{k+1}{e}^{-2x})$

So the statement is also true for $n=k+1$.

Now by principle of mathematical induction we have the statement is true $\forall n\in N$.