91Ó°ÊÓ

We are given the Maclaurin series of$\sin x,\cos x,e^x$

The Maclaurin series of $\sin x$can be given as

role="math" localid="1653924929632" $$\begin{gathered} \sin x=x-\frac{x^3}{6}+\frac{x^5}{5!}-...... \\ \sin x=\underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^n}{(2n+1)!}{x}^{2n+1} \end{gathered}$$

Differentiating term by term

we get,

$$\begin{gathered} \frac{d\left(\sin x\right)}{dx}=\frac{d\underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^n}{(2n+1)!}{x}^{2n+1}}{dx} \\ \frac{d\left(\sin x\right)}{dx}=\underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^n}{\left(2n\right)!}{x}^{2n} \\ but\underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^n}{\left(2n\right)!}{x}^{2n}=\cos x \\ Therefore,\frac{d\left(\sin x\right)}{dx}=\cos x \end{gathered}$$

The Maclaurin series of cos x can be given as

$\cos x=\underset{k=0}{\overset{∞}{∑}}\frac{{(-1)}^k{x}^{2k}}{\left(2k\right)!}$

Differentiating term by term we get,

$$\begin{gathered} \frac{d\left(\cos x\right)}{dx}=\frac{d\left(\underset{k=0}{\overset{∞}{∑}}\frac{{(-1)}^k{x}^{2k}}{\left(2k\right)!}\right)}{dx} \\ \frac{d\left(\cos x\right)}{dx}=\underset{k=1}{\overset{∞}{∑}}\frac{{(-1)}^k{x}^{2k-1}}{(2k-1)!} \\ \frac{d\left(\cos x\right)}{dx}=\underset{k=1}{\overset{∞}{-∑}}\frac{{(-1)}^k{x}^{2k-1}}{(2k-1)!} \\ therefore \\ \frac{d\left(\cos x\right)}{dx}=-\sin x \end{gathered}$$

The Maclaurin series of $e^x$is given by

$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+.....$

Differentiating term by term

We get,

$$\begin{gathered} \frac{d{e}^x}{dx}=\frac{d(1+x+{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^3}{3!}})}{dx} \\ \frac{d{e}^x}{dx}=1+x+\frac{x^2}{2}+\frac{x^3}{3!} \\ \frac{d{e}^x}{dx}=e^x \end{gathered}$$