Problem 50 - Thomas Calculus

Question

Derive the formula for the derivative with respect to \(x\) of $$ \text { a. sec }x . \quad \text { b. csc } x . \quad \text { c. } \cot x $$

Step-by-Step Solution

Answer

a. \( \sec(x) \tan(x) \), b. \( -\csc(x) \cot(x) \), c. \( -\csc^2(x) \).

Step 1: Derivative of secant function

The function to differentiate is \( \sec(x) \). We know that \( \sec(x) = \frac{1}{\cos(x)} \). Using the quotient rule, where \( u(x) = 1 \) and \( v(x) = \cos(x) \), we find \( u'(x) = 0 \) and \( v'(x) = -\sin(x) \). The quotient rule states that \( \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). Substituting in gives:\[ \frac{d}{dx}\left( \sec(x) \right) = \frac{0 \cdot \cos(x) - 1 \cdot (-\sin(x))}{\cos^2(x)} = \frac{\sin(x)}{\cos^2(x)}. \]This simplifies using \( \tan(x) = \frac{\sin(x)}{\cos(x)} \) and \( \sec(x) = \frac{1}{\cos(x)} \) to:\[ \frac{d}{dx} \sec(x) = \sec(x) \tan(x). \]

Step 2: Derivative of cosecant function

The function to differentiate is \( \csc(x) \). We know that \( \csc(x) = \frac{1}{\sin(x)} \). Using the quotient rule with \( u(x) = 1 \) and \( v(x) = \sin(x) \), gives \( u'(x) = 0 \) and \( v'(x) = \cos(x) \). Substitute into the quotient rule:\[ \frac{d}{dx}\left( \csc(x) \right) = \frac{0 \cdot \sin(x) - 1 \cdot \cos(x)}{\sin^2(x)} = -\frac{\cos(x)}{\sin^2(x)}. \]This simplifies using \( \cot(x) = \frac{\cos(x)}{\sin(x)} \) and \( \csc(x) = \frac{1}{\sin(x)} \) to:\[ \frac{d}{dx} \csc(x) = -\csc(x) \cot(x). \]

Step 3: Derivative of cotangent function

The function to differentiate is \( \cot(x) \). We know that \( \cot(x) = \frac{\cos(x)}{\sin(x)} \). Using the quotient rule with \( u(x) = \cos(x) \) and \( v(x) = \sin(x) \), we have \( u'(x) = -\sin(x) \) and \( v'(x) = \cos(x) \). Applying the quotient rule:\[ \frac{d}{dx}\left( \cot(x) \right) = \frac{-\sin(x) \cdot \sin(x) - \cos(x) \cdot \cos(x)}{\sin^2(x)} = \frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)}. \]Since \( \sin^2(x) + \cos^2(x) = 1 \), this simplifies to:\[ \frac{d}{dx} \cot(x) = -\frac{1}{\sin^2(x)} = -\csc^2(x). \]

Key Concepts

Derivative of Secant Function

To find the derivative of the secant function, given as \( \sec(x) \), we first recognize that \( \sec(x) \) can be expressed in terms of cosine: \( \sec(x) = \frac{1}{\cos(x)} \).
We use the quotient rule, which is a fundamental tool in calculus for differentiating functions that are expressed as a quotient of two other functions.
Here, we set \( u(x) = 1 \) and \( v(x) = \cos(x) \).

• Derivative of the numerator: \( u'(x) = 0 \)
• Derivative of the denominator: \( v'(x) = -\sin(x) \)

Substituting these into the quotient rule formula \( \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \), we get:\[\frac{d}{dx}\left( \sec(x) \right) = \frac{0 \cdot \cos(x) - 1 \cdot (-\sin(x))}{\cos^2(x)} = \frac{\sin(x)}{\cos^2(x)}.\]To simplify further, we use trigonometric identities: \( \tan(x) = \frac{\sin(x)}{\cos(x)} \) and \( \sec(x) = \frac{1}{\cos(x)} \). Thus, the derivative becomes:\[\frac{d}{dx} \sec(x) = \sec(x) \tan(x).\] The result shows that the rate of change of \( \sec(x) \) is directly related to both itself and \( \tan(x) \). This results from the way secant stretches and elevates the value of the tangent across angles.

Derivative of Cosecant Function

The derivative of the cosecant function, \( \csc(x) \), requires a similar approach to that of the secant function.
We know that \( \csc(x) = \frac{1}{\sin(x)} \).
Applying the quotient rule again with \( u(x) = 1 \) and \( v(x) = \sin(x) \), we obtain:

• Derivative of the numerator: \( u'(x) = 0 \)
• Derivative of the denominator: \( v'(x) = \cos(x) \)

Substituting these into the quotient rule, we find:\[\frac{d}{dx}\left( \csc(x) \right) = \frac{0 \cdot \sin(x) - 1 \cdot \cos(x)}{\sin^2(x)} = -\frac{\cos(x)}{\sin^2(x)}.\]To simplify this further, we use the cofunction \( \cot(x) = \frac{\cos(x)}{\sin(x)} \) and \( \csc(x) = \frac{1}{\sin(x)} \).
Thus, the expression simplifies to:\[\frac{d}{dx} \csc(x) = -\csc(x) \cot(x).\] This derivative shows how the rate of change of \( \csc(x) \) is impacted by both cosecant and cotangent functions, reflecting their interrelated nature on the unit circle.

Derivative of Cotangent Function

Lastly, understanding the derivative of the cotangent function, \( \cot(x) \), is crucial.
Expressing \( \cot(x) \) as a quotient of trigonometric functions, we have \( \cot(x) = \frac{\cos(x)}{\sin(x)} \).
Using the quotient rule, we identify:

• Numerator: \( u(x) = \cos(x) \), so \( u'(x) = -\sin(x) \)
• Denominator: \( v(x) = \sin(x) \), so \( v'(x) = \cos(x) \)

Applying the quotient rule:\[\frac{d}{dx}\left( \cot(x) \right) = \frac{-\sin(x) \cdot \sin(x) - \cos(x) \cdot \cos(x)}{\sin^2(x)}.\]Utilizing the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \), this simplifies to:\[\frac{d}{dx} \cot(x) = -\frac{1}{\sin^2(x)} = -\csc^2(x).\] This derivative highlights the strong link between cotangent and cosecant functions, where \( \cot(x) \) naturally transforms into a negative square of \( \csc(x) \), emphasizing the intrinsic properties and relationships within trigonometric identities.