/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 Prove the following differentiat... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 67

Prove the following differentiation rules. (a) \(\frac{d}{d x}[\sec x]=\sec x \tan x\) (b) \(\frac{d}{d x}[\csc x]=-\csc x \cot x\) (c) \(\frac{d}{d x}[\cot x]=-\csc ^{2} x\)

Short Answer

Expert verified
The derivative of \(\sec x\) is \(\sec x \tan x\), the derivative of \(\csc x\) is \(-\csc x \cot x\), and the derivative of \(\cot x\) is \(-\csc^{2} x\). Use the quotient rule for calculus and trigonometric identities to prove this.

Step by step solution

01

Prove the derivative of sec x

The sec function can be rewritten as \( \sec x = \frac{1}{\cos x} \). Use the quotient rule for derivatives: \(\frac{d}{d x} [f/g] = \frac{g f' - f g'}{g^2}\). This results in \(\frac{d}{d x} [\sec x] = \frac{d}{d x} [1 / \cos x] = \frac{\cos x * 0 - 1 * (-\sin x)}{(\cos x)^2} = \frac{\sin x}{(\cos x)^2} = \sec x \tan x. \)
02

Prove the derivative of csc x

The csc function can be rewritten as \( \csc x = \frac{1}{\sin x} \). Using the quotient rule for derivatives, we get \(\frac{d}{d x} [\csc x] = \frac{d}{d x} [1 / \sin x] = \frac{\sin x * 0 - 1 * (\cos x)}{(\sin x)^2} = -\frac{\cos x}{(\sin x)^2} = -\csc x \cot x.\)
03

Prove the derivative of cot x

Use the identity \( \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x} \). Again, use the quotient rule for derivatives to get \(\frac{d}{d x} [\cot x] = \frac{d}{d x} [\cos x / \sin x] = \frac{\sin x * (-\sin x) - \cos x * \cos x}{(\sin x)^2} = -\frac{1}{(\sin x)^2} = -\csc ^{2} x.\)

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