Problem 205 - Calculus Volume 1

Question

Use the quotient rule to derive the given equations. $$ \frac{d}{d x}(\cot x)=-\csc ^{2} x $$

Step-by-Step Solution

Answer

The derivative of \( \cot x \) is \(-\csc^2 x \).

Step 1: Identify the function to be derived

Our given function is \( \cot x \), which can be rewritten as \( \frac{\cos x}{\sin x} \) because \( \cot x = \frac{\cos x}{\sin x} \). We will use this form to apply the quotient rule.

Step 2: Apply the Quotient Rule

The quotient rule states that if \( u(x) \) and \( v(x) \) are differentiable functions, then the derivative of \( \frac{u}{v} \) is given by \( \frac{u'v - uv'}{v^2} \). Set \( u(x) = \cos x \) and \( v(x) = \sin x \). Now compute \( u' = -\sin x \) and \( v' = \cos x \).

Step 3: Compute the Numerator of the Quotient Rule

Substitute into the numerator: \( u'v - uv' = (-\sin x)(\sin x) - (\cos x)(\cos x) \). This simplifies to \(-\sin^2 x - \cos^2 x \).

Step 4: Simplify the Numerator Using Trigonometric Identities

Use the identity \( \sin^2 x + \cos^2 x = 1 \) to rewrite \(-\sin^2 x - \cos^2 x = -(\sin^2 x + \cos^2 x) = -1 \).

Step 5: Compute the Denominator of the Quotient Rule

The denominator is simply \( v^2 = (\sin x)^2 = \sin^2 x \).

Step 6: Combine the Results to Find the Derivative

Put the simplified numerator and denominator together: \( \frac{d}{dx}(\cot x) = \frac{-1}{\sin^2 x} \). This simplifies to \(-\csc^2 x \) because \( \csc x = \frac{1}{\sin x} \).

Step 7: State the Result

The derivative using the quotient rule confirms the result: \( \frac{d}{d x}(\cot x) = -\csc^2 x \).

Key Concepts

Quotient Rule

The Quotient Rule is a method in calculus for differentiating functions that are expressed as a quotient, which means one function divided by another. This technique is essential when you have a function of the form \( \frac{u(x)}{v(x)} \). The rule itself states that the derivative of such a function is \( \frac{u'v - uv'}{v^2} \).
Here's a simple breakdown:

• Identify your numerator \( u(x) \) and denominator \( v(x) \).
• Calculate the derivative of the numerator \( u'(x) \).
• Calculate the derivative of the denominator \( v'(x) \).
• Plug these into the formula: \( u'v - uv' \) over \( v^2 \).

So for our exercise with \( \cot x \), consider \( u(x) = \cos x \) and \( v(x) = \sin x \), then apply the rule. Computation involves substituting and simplification using known identities like \( \sin^2 x + \cos^2 x = 1 \). The key here is to handle expressions accurately, ensuring the result is correct, like getting \(-\csc^2 x\).

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, involving functions of an angle. These functions relate the angles of a triangle to the lengths of its sides. In calculus, they often appear in differentiation problems.
The standard trigonometric functions include:

• \( \sin x \): Sine of angle \( x \)
• \( \cos x \): Cosine of angle \( x \)
• \( \tan x \): Tangent of angle \( x \)
• \( \cot x \): Cotangent, defined as \( \frac{\cos x}{\sin x} \)
• \( \sec x \): Secant, the reciprocal of cosine
• \( \csc x \): Cosecant, the reciprocal of sine

When differentiating, knowing how each function behaves can simplify complex calculations. For instance, the derivative of \( \sin x \) is \( \cos x \), and for \( \cos x \), it is \(-\sin x \). This relationship is pivotal in solving problems involving trigonometric derivatives.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold for all values of the involved variables where both sides are defined. These identities are crucial in both simplifying expressions and solving equations, particularly when differentiating trigonometric functions.
Key identities include:

• Pythagorean Identity: \( \sin^2 x + \cos^2 x = 1 \).
• Reciprocal Identities: \( \csc x = \frac{1}{\sin x} \) and \( \sec x = \frac{1}{\cos x} \).
• Quotient Identities: \( \tan x = \frac{\sin x}{\cos x} \) and \( \cot x = \frac{\cos x}{\sin x} \).

These identities simplify differentiation tasks, as they allow replacement of one expression with another equivalent expression to make derivatives easier to compute. In our exercise, using \( \sin^2 x + \cos^2 x = 1 \), we transformed the complex expression \(-\sin^2 x - \cos^2 x\) into \(-1\), significantly simplifying the problem by minimizing the number of terms.