Q. 24 - Calculus

Question

(a) Verify that$\frac{d}{dx}\left(x\sin \left(x^3\right)\right)=\sin \left(x^3\right)+3x^3\cos \left(x^3\right)$

(b) Use multiplication and/or substitution in the Maclaurin series for the sine and the cosine to find the Maclaurin series for $x\sin \left(x^3\right)$,$\sin \left(x^3\right)$ and $3x^3\cos \left(x^3\right)$

(c) Use Theorem 8.11 to find the Maclaurin series for $\frac{d}{dx}\left(x\sin \left(x^3\right)\right)$, and show that this series is the sum of the Maclaurin series for $\sin \left(x^3\right)$ and $3x^3\cos \left(x^3\right)$ you obtained in part (b).

Step-by-Step Solution

Answer

(a) The $\frac{d}{dx}\left(x\sin \left(x^3\right)\right)=\sin \left(x^3\right)+3x^3\cos \left(x^3\right)$ is true.

(b) The Maclaurin series is $x\sin \left(x^3\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{6k+4}$.

(c) The sum of the Maclaurin series is  $3x^3\cos \left(x^3\right)=3\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{6k+3}$.

Part (a) : Step 1 : Given information

Given function : $\frac{d}{dx}\left(x\sin \left(x^3\right)\right)=\sin \left(x^3\right)+3x^3\cos \left(x^3\right)$

Part (a) Step 2 : Simplification

Equation :$\frac{d}{dx}\left(x\sin \left(x^3\right)\right)=\sin \left(x^3\right)+3x^3\cos \left(x^3\right)$,

Let's begin with the left side, which we'll solve by separating$x\sin \left(x^3\right)$ with respect to $x$.

Therefore,

$\begin{array}{rcl}\frac{d}{dx}\left(x\sin \left(x^3\right)\right)& =& x\frac{d}{dx}\left[\sin \left(x^3\right)\right]+\sin \left(x^3\right)\frac{d}{dx}\left[x\right]\\ & =& x\cos \left(x^3\right)·3x^2+\sin \left(x^3\right)·1\\ & =& 3x^3\cos \left(x^3\right)+\sin \left(x^3\right)\end{array}$

Hence proved that $\frac{d}{dx}\left(x\sin \left(x^3\right)\right)=\sin \left(x^3\right)+3x^3\cos \left(x^3\right)$

Part (b): Step 1 : Given information

Given functions : $x\sin \left(x^3\right) , \sin \left(x^3\right) \mathrm{and} 3x^3\cos \left(x^3\right)$

Part (b): Step 2 : Simplification

For function$\sin x$ , the Maclaurin series is

$\sin x=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{2k+1}$

So, in the Maclaurin series of $\sin x$, replace $x$ with $x^3$ to discover the Maclaurin series for the function .

$$\begin{gathered} \sin \left(x^3\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{\left(x^3\right)}^{2k+1} \\ \Rightarrow \sin \left(x^3\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{6k+3} \end{gathered}$$

Also, the Maclaurin series for $x\sin \left(x^3\right)$ can be found by multiplying $x$ with the Maclaurin series of $\sin \left(x^3\right)$

So,

$$\begin{gathered} x\sin \left(x^3\right)=x\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{6k+3} \\ \Rightarrow \mathit{x}\mathit{s}\mathit{i}\mathit{n}\left(x^3\right)\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{k}}}{\mathbf{(}\mathbf{2}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{!}}{\mathit{x}}^{\mathbf{6}\mathbf{k}\mathbf{+}\mathbf{4}} \end{gathered}$$

Part (c): Step 1 : Given information

Given function : $\frac{d}{dx}\left(x\sin \left(x^3\right)\right)$

Part (c): Step 2 : Simplification

Similarly, for the function $\cos x$ the Maclaurin series is

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

In the Maclaurin series of $\cos x$, replace $x$ with $x^3$ to discover the Maclaurin series for the function.

It means that

$\begin{array}{rcl}\cos \left(x^3\right)& =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{\left(x^3\right)}^{2k}\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{6k}\end{array}$

The Maclaurin series for $3x^3\cos \left(x^3\right)$ can be found by multiplying $x$ with the Maclaurin series of $\cos \left(x^3\right)$

So,

$$\begin{gathered} 3x^3\cos \left(x^3\right)=3x^3\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{6k} \\ \Rightarrow \mathbf{3}{\mathit{x}}^{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}{\mathbf{x}}^{\mathbf{3}}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{k}}}{\mathbf{\left(}\mathbf{2}\mathbf{k}\mathbf{\right)}\mathbf{!}}{\mathit{x}}^{\mathbf{6}\mathbf{k}\mathbf{+}\mathbf{3}} \end{gathered}$$