Chapter 10: Q. 1 (page 777)

Use the Maclaurin series for $\sin (x), \cos (x), e^{x}$ to prove that
role="math" localid="1653925824016" $a) \frac{d (\sin x)}{d x} = \cos x b) \frac{d (\cos x)}{d x} = - \sin x c) \frac{d (e^{x})}{d x} = e^{x}$

Short Answer

Expert verified

We proved that

$a) \frac{d (\sin x)}{d x} = \cos x b) \frac{d (\cos x)}{d x} = - \sin x c) \frac{d (e^{x})}{d x} = e^{x}$

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" $\sin x = x - \frac{x^{3}}{6} + \frac{x^{5}}{5!} - . . . . . . \sin x = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} x^{2 n + 1}$

Differentiating term by term

we get,

$\frac{d (\sin x)}{d x} = \frac{d \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} x^{2 n + 1}}{d x} \frac{d (\sin x)}{d x} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} x^{2 n} b u t \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} x^{2 n} = \cos x T h e r e f o r e, \frac{d (\sin x)}{d x} = \cos x$

Part b) Step 1: Proof

The Maclaurin series of cos x can be given as

$\cos x = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} x^{2 k}}{(2 k)!}$

Differentiating term by term we get,

$\frac{d (\cos x)}{d x} = \frac{d (\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} x^{2 k}}{(2 k)!})}{d x} \frac{d (\cos x)}{d x} = \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k} x^{2 k - 1}}{(2 k - 1)!} \frac{d (\cos x)}{d x} = {- \sum}_{k = 1}^{\infty} \frac{{(- 1)}^{k} x^{2 k - 1}}{(2 k - 1)!} t h e r e f o r e \frac{d (\cos x)}{d x} = - \sin x$

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,

$\frac{d e^{x}}{d x} = \frac{d (1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3!})}{d x} \frac{d e^{x}}{d x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3!} \frac{d e^{x}}{d x} = e^{x}$

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Use the Maclaurin series for $\sin (x), \cos (x), e^{x}$ to prove that
role="math" localid="1653925824016" $a) \frac{d (\sin x)}{d x} = \cos x b) \frac{d (\cos x)}{d x} = - \sin x c) \frac{d (e^{x})}{d x} = e^{x}$

Short Answer

Step by step solution

Given information

Part a) Step 1: Proof

Part b) Step 1: Proof

Part c) Step 1: Proof

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Use the Maclaurin series for sinx,cosx,exto prove thatrole="math" localid="1653925824016" a)d(sinx)dx=cosxb)d(cosx)dx=-sinxc)d(ex)dx=ex

Short Answer

Step by step solution

Given information

Part a) Step 1: Proof

Part b) Step 1: Proof

Part c) Step 1: Proof

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Use the Maclaurin series for $\sin (x), \cos (x), e^{x}$ to prove that
role="math" localid="1653925824016" $a) \frac{d (\sin x)}{d x} = \cos x b) \frac{d (\cos x)}{d x} = - \sin x c) \frac{d (e^{x})}{d x} = e^{x}$