Problem 8 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$g(x)=\frac{\cos x}{\sin ^{2} x}$$

Step-by-Step Solution

Answer

The derivative is \(-\frac{\sin x + 2\cos^2 x \sin x}{\sin^4 x}\)."

Step 1: Identify the Function

The given function is \( g(x) = \frac{\cos x}{\sin^2 x} \). We need to differentiate this function with respect to \( x \). This function is a quotient of two functions: \( \cos x \) (numerator) and \( \sin^2 x \) (denominator).

Step 2: Apply the Quotient Rule

The quotient rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of \( \frac{u}{v} \) is \( \frac{v \cdot u' - u \cdot v'}{v^2} \). Here, \( u(x) = \cos x \) and \( v(x) = \sin^2 x \).

Step 3: Differentiate the Numerator

Differentiate \( u(x) = \cos x \) to get \( u'(x) = -\sin x \). This derivative is derived using the standard derivative of \( \cos x \), which is \(-\sin x \).

Step 4: Differentiate the Denominator

Differentiating the denominator \( v(x) = \sin^2 x \) requires the use of the chain rule. Rewrite \( \sin^2 x \) as \((\sin x)^2 \), and apply the chain rule to get \( 2\sin(x)\cos(x) \).

Step 5: Substitute into the Quotient Rule Formula

Substitute the derivatives into the quotient rule formula: \[\frac{d}{dx}\left(\frac{\cos x}{\sin^2 x}\right) = \frac{\sin^2 x \cdot (-\sin x) - \cos x \cdot 2\sin x\cos x}{(\sin^2 x)^2}.\] Simplifying the numerator gives \(-\sin^3 x - 2\cos^2 x \sin x\).

Step 6: Simplify the Derivative Expression

Factor out \( -\sin x \) from the numerator: \(-\sin x (\sin^2 x + 2 \cos^2 x) \). Using \( \sin^2 x + \cos^2 x = 1 \) in the expression, the derivative becomes: \[-\sin x (1 + \cos^2 x) = -\sin x - 2\cos^2 x\sin x.\]

Step 7: Write the Final Expression

The expression of the derivative becomes: \[\frac{-\sin x (1 + \cos^2 x)}{\sin^4 x} = -\frac{\sin x + 2\cos^2 x \sin x}{\sin^4 x}.\] This simplifies to \(-\frac{\sin^3 x + 2\cos^2 x \sin x}{\sin^4 x}\).

Step 8: Cancel Common Factors (if any)

If possible, check for common factors and further simplify. In this case, the expression is already simplest: \(-\frac{\sin x + 2\cos^2 x \sin x}{\sin^4 x}\).

Step 9: Final Answer

The derivative of the function \( g(x) = \frac{\cos x}{\sin^2 x} \) is \(-\frac{\sin x + 2\cos^2 x \sin x}{\sin^4 x}\).

Key Concepts

Quotient Rule

When you come across a function that is a fraction, where one function is divided by another, you can use the quotient rule to find the derivative efficiently. In this problem, we have the function \( g(x) = \frac{\cos x}{\sin^2 x} \). Here our numerator \( u(x) \) is \( \cos x \) and the denominator \( v(x) \) is \( \sin^2 x \).

• The quotient rule is given by the formula:\[\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}\]
• To apply the quotient rule, find the derivatives of both the numerator and the denominator. This is crucial because it forms the basis of substituting into the formula.

It's important to follow this order: derive the components first, then substitute them correctly. In this exercise, remember to simplify your work by finding possible common factors in the last step.

Chain Rule

The chain rule is a way to differentiate more complex expressions that involve a composition of functions. In the context of this exercise, when differentiating the denominator \( v(x) = \sin^2 x \), we must apply the chain rule. This is because \( \sin^2 x \) is essentially a composition of two functions: the expression \((\sin x)^2\).
Here's how to think about it:

• Identify the inner function as \( \sin x \) and the outer function as \( x^2 \).
• Differentiate the outer function with respect to the inner function: \( 2x \rightarrow 2\sin x \).
• Differentiate the inner function: \( \cos x \).
• The derivative of \( v(x) \) becomes \( 2\sin x \cdot \cos x \).

Applying the chain rule correctly ensures you can tackle more complex derivatives involving nested functions.

Trigonometric Functions

Trigonometric functions often appear in calculus problems, and understanding their derivatives is key. In this problem, we deal with trig functions \( \cos x \) and \( \sin x \).

• The derivative of \( \cos x \) is \( -\sin x \).
• For \( \sin x \), the derivative is \( \cos x \).

These fundamental derivatives are used when applying both the quotient rule and the chain rule. They complement each other as they are frequently used to build and simplify equations in calculus. Also, the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) helps simplify expressions, as seen in changing \( \sin^2 x + 2 \cos^2 x \) during simplification. Knowing these relationships helps make complex expressions more manageable.