Question

Use the Maclaurin series for $\sin \left(x\right),\cos \left(x\right),e^x$to prove that

role="math" localid="1653925824016" $$\begin{gathered} a)\frac{d\left(\sin x\right)}{dx}=\cos x \\ b)\frac{d\left(\cos x\right)}{dx}=-\sin x \\ c)\frac{d\left(e^x\right)}{dx}=e^x \end{gathered}$$

Short answer

We proved that

$$\begin{gathered} a)\frac{d\left(\sin x\right)}{dx}=\cos x \\ b)\frac{d\left(\cos x\right)}{dx}=-\sin x \\ c)\frac{d\left(e^x\right)}{dx}=e^x \end{gathered}$$

Step-by-step solution

Given information

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

Part a) Step 1: Proof

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

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

Differentiating term by term

we get,

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

Part b) Step 1: Proof

The Maclaurin series of cos x can be given as

$\cos x=\sum _{k=0}^{\infty }\frac{{(-1)}^k{x}^{2k}}{\left(2k\right)!}$

Differentiating term by term we get,

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

Part c) Step 1: Proof

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,

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