Question

Proof. Prove the following differentiation rules. $$ \begin{array}{l}{\text { (a) } \frac{d}{d x}[\sec x]=\sec x \tan x} \\ {\text { (b) } \frac{d}{d x}[\csc x]=-\csc x \cot x} \\ {\text { (c) } \frac{d}{d x}[\cot x]=-\csc ^{2} x}\end{array} $$

Short answer

The proofs for the derivatives of secant, cosecant, and cotangent are as follow: (a) The derivative of secant x is secant x times tangent x. (b) The derivative of cosecant x is negative cosecant x times cotangent x. (c) The derivative of cotangent x is negative cosecant square x.

Step-by-step solution

Deriving Rule (a)

Firstly, remember that \(\sec x = \frac{1}{\cos x}\). Applying the quotient rule for differentiation, in this case, where the denominator is a function of x gives us \(\frac{d}{dx} [\sec x] = \frac{-1 \cdot -\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \sec x \tan x\)

Deriving Rule (b)

Recall the identity \(\csc x = \frac{1}{\sin x}\). Use the quotient rule once more to find the derivative: \( \frac{d}{dx} [\csc x] = \frac{-1 \cdot \cos x}{\sin^2 x} = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = -\csc x \cot x\)

Deriving Rule (c)

Another identity to recall is that \(\cot x = \frac{\cos x}{\sin x}\). Again apply the quotient rule to find the derivative, which will yield: \( \frac{d}{dx} [\cot x] = \frac{\sin x \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x} = \frac{1 - \sin^2 x}{\sin^2 x}\) . Using the Pythagorean identity, \( 1 - \sin^2 x = \cos^2 x\) , we arrive at \(\frac{d}{dx} [\cot x] = -\frac{\cos^2 x}{\sin^2 x} = -\csc^2 x\)

Key concepts

Quotient Rule for Differentiation

Understanding the quotient rule for differentiation is a key aspect of calculus, especially when dealing with complex functions. In essence, this rule allows us to find the derivative of a function that is the quotient of two differentiable functions. The formula for the quotient rule is given by:
\
\[\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2}\]
When applying this technique, it is important to identify the top function as \(u(x)\) and the bottom function as \(v(x)\), take their derivatives separately \(u'(x)\) and \(v'(x)\), and then plug these values into the formula. It's helpful to keep track of the signs, especially the subtraction in the numerator, as a small mistake can lead to the wrong answer.

Trigonometric Functions Derivatives

The derivatives of trigonometric functions are fundamental in calculus, and learning them can make complex differentiation problems easier to tackle. Each trigonometric function—sine, cosine, tangent, secant, cosecant, and cotangent—has a specific derivative:

• The derivative of \(\sin x\) is \(\cos x\).
• The derivative of \(\cos x\) is \( -\sin x\).
• The derivative of \(\tan x\) is \(\sec^2 x\).

These derivations are often memorized, but understanding how they are derived, for instance, through the limit definition of a derivative, reinforces comprehension. Leveraging these derivatives simplifies the process of finding the derivatives of more complex trigonometric expressions.

Pythagorean Identity

The Pythagorean identity is an essential tool in calculus that relates the squares of the sine and cosine functions to 1:
\
\[\sin^2 x + \cos^2 x = 1\]
This identity is extremely useful when simplifying expressions during differentiation and integration, as it allows us to convert between \(\sin^2 x\) and \(\cos^2 x\). For instance, if we need \(\cos^2 x\) but we have \(\sin^2 x\), we can simply use the identity to rewrite \(\sin^2 x\) as \(1 - \cos^2 x\), or vice versa. This conversion is crucial for solving many trigonometric differentiation problems.

Secant Derivative

When it comes to differentiating the secant function, the result involves another trigonometric function:
\
\[\frac{d}{dx}[\sec x] = \sec x \tan x\]
The derivative of \(\sec x\), which is \(\frac{1}{\cos x}\), can be found by using the quotient rule where \(u(x) = 1\) and \(v(x) = \cos x\), as showcased in the exercise solution. Remember that \(\tan x = \frac{\sin x}{\cos x}\), which simplifies the expression to the secant multiplied by the tangent of x. Recognizing these relationships helps significantly in solving derivatives of trigonometric functions.

Cosecant Derivative

Similar to the secant function, the derivative of the cosecant function, \(\csc x = \frac{1}{\sin x}\), also involves another trigonometric function:
\
\[\frac{d}{dx}[\csc x] = -\csc x \cot x\]
By using the quotient rule with \(u(x) = 1\) and \(v(x) = \sin x\), the derivative of the cosecant incorporates the negative sign, which is important to note. The function \(\cot x\) here represents \(\frac{\cos x}{\sin x}\), and hence, the negative cosecant function is multiplied by the cotangent function. These interrelations are crucial for understanding how to derive trigonometric functions.

Cotangent Derivative

The derivative of the cotangent function is unique among the basic trigonometric functions:
\
\[\frac{d}{dx}[\cot x] = -\csc^2 x\]
To derive this using the quotient rule, consider \(u(x) = \cos x\) and \(v(x) = \sin x\), then apply the rule. The use of the Pythagorean identity converts \(1 - \sin^2 x\) to \(\cos^2 x\), and as a result, we get the negative cosecant squared. Remembering this distinct result, which doesn't involve another trigonometric function directly, can be helpful while working through differentiation problems that include cotangent functions.