Problem 12 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$y=\frac{\cos x}{1+\sin x}$$

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{dx} = \frac{-\sin x - 1}{(1 + \sin x)^2}\).

Step 1: Identify the Rule for Differentiation

We have the function \(y = \frac{\cos x}{1 + \sin x}\). This is a quotient, so we will use the Quotient Rule for differentiation, which states that if \(y = \frac{u(x)}{v(x)}\), then \(\frac{dy}{dx} = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}\). Here, \(u(x) = \cos x\) and \(v(x) = 1 + \sin x\).

Step 2: Differentiate the Numerator (u(x))

The numerator of the function is \(u(x) = \cos x\). The derivative \(u'(x)\) is given by \(-\sin x\).

Step 3: Differentiate the Denominator (v(x))

The denominator of the function is \(v(x) = 1 + \sin x\). The derivative \(v'(x)\) is \(\cos x\).

Step 4: Apply the Quotient Rule

Substitute \(u(x)\), \(u'(x)\), \(v(x)\), and \(v'(x)\) into the Quotient Rule formula. Thus, we have:\[\frac{dy}{dx} = \frac{(1 + \sin x)(-\sin x) - \cos x(\cos x)}{(1 + \sin x)^2}\]

Step 5: Simplify the Expression

Simplify the expression:\[\frac{dy}{dx} = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2}\]We know \(\sin^2 x + \cos^2 x = 1\), so the expression becomes:\[\frac{dy}{dx} = \frac{-\sin x - 1}{(1 + \sin x)^2}\]

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. In a more intuitive sense, it tells us the rate of change or the slope of the function at any given point. For a function \( y = f(x) \), the derivative is denoted as \( \frac{dy}{dx} \) and provides information about the function's behavior, such as whether it is increasing or decreasing.
For a quotient of two functions, such as \( y = \frac{u(x)}{v(x)} \), the Quotient Rule is a specific technique used to differentiate this kind of function. The rule formula is \( \frac{dy}{dx} = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \), where \( u(x) \) and \( v(x) \) are differentiable functions.
- **Key Points**

• The numerator \( u(x) \) and the denominator \( v(x) \) are both functions of \( x \).
• The derivative of the quotient depends on the derivatives of both \( u(x) \) and \( v(x) \).
• This rule is essential when dealing with ratios of functions in calculus.

Trigonometric Functions

Trigonometric functions are functions of an angle and are important in mathematics, particularly in calculus. Common trigonometric functions include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)). These functions, which relate angles to ratios of side lengths in right triangles, are periodic and have derivatives that play a key role in calculus.
For example: - The derivative of \( \sin x \) is \( \cos x \).- The derivative of \( \cos x \) is \( -\sin x \).
These derivatives are vital when applying the Quotient Rule in functions that involve trigonometric expressions.
Considering our problem \( y = \frac{\cos x}{1 + \sin x} \), the numerator \( \cos x \) is differentiated using its derivative rule, resulting in \( -\sin x \). The denominator \( 1 + \sin x \), when differentiated, uses the derivative of sine, yielding \( \cos x \).
- **Key Characteristics**

• Trigonometric derivatives often involve the other trigonometric functions.
• They are periodic, allowing for similar behavior of the derived function over a period.
• Understanding their derivatives helps solve complex calculus problems that involve waves and oscillations.

Calculus

Calculus is a branch of mathematics that studies continuous change. It is divided into differential calculus and integral calculus. Differential calculus focuses on the concept of the derivative, which measures change, while integral calculus concerns accumulation of quantities.
The exercise provides a practical application of differential calculus through the Quotient Rule. By differentiating the function \( y = \frac{\cos x}{1 + \sin x} \), we apply differential calculus to find the rate at which one trigonometric function changes with respect to another.
In practical terms:

• Differential calculus helps in understanding the slope and behavior of graphs.
• It is crucial in physics for understanding dynamics and motion.
• Financial analysts utilize calculus to model economic predictions based on changing variables.

Calculus allows us to simplify complex rate-of-change problems into manageable calculations, like differentiating the given function using trigonometric identities. Remember that by knowing \( \sin^2 x + \cos^2 x = 1 \), we were able to simplify the derivative result to \( \frac{-\sin x - 1}{(1 + \sin x)^2} \). This highlights calculus's power in providing insights through mathematical simplifications.