Question

Assuming that \((d / d x)(\sin x)=\cos x\) and \((d / d x)(\cos x)=\) \(-\sin x,\) prove each of the following. (a) \(\frac{d}{d x} \tan x=\sec ^{2} x\) (b)\(\frac{d}{d x} \sec x=\sec x \tan x\)

Short answer

(a) \( \frac{d}{d x} \tan x = \sec ^{2} x \) (b) \( \frac{d}{d x} \sec x = \sec x \tan x \)

Step-by-step solution

Differentiate \( \tan x \)

The expression \( \tan x \) can be rewritten as \( \frac{\sin x}{\cos x} \). To differentiate \( \tan x \), the quotient rule can be used, which is \( (f/g)' = (g*f' - f*g') / g^2 \). Let \( f = \sin x \) and \( g = \cos x \). Substituting these into the formula, the next step becomes the differentiation of \( f \) and \( g \), which according to the given information returns \( \cos x \) and \( -\sin x \) respectively. Thus, we have \( \frac{d}{dx} \tan x = \frac{\cos^2x + \sin^2x}{\cos^2x} \).

Simplify using the Pythagorean Identity

Using the Pythagorean trigonometric identity \( \sin^2x + \cos^2x = 1 \), the equation simplifies to \( \frac{d}{dx} \tan x = \frac{1}{\cos^2x} \), which is equivalent to \( \sec^2x \). Thus, the derivative of \( \tan x \) is \( \sec^2x \).

Differentiate \( \sec x \)

The expression \( \sec x \) is the reciprocal of \( \cos x \), thus, can be written as \( 1/ \cos x \). To find the derivative of \( \sec x \), the quotient rule needs to be applied again just like in step 1. By using the same formula, the next step is the differentiation and simplification of the obtained equation.

Simplify the differentiation of \( \sec x \)

After differentiating \( 1/ \cos x \) with respect to \( x \) using quotient rule, we obtain \( \frac{d}{dx} \sec x = \frac{\sin x}{\cos^2 x} \). Now using trigonometric identities and simplifying, \( \frac{\sin x}{\cos x} * \frac{1}{\cos x} \) turns into \( \sec x \tan x \). Thus, the derivative of \( \sec x \) is \( \sec x \tan x \).