Question

Find \(dy/dx\) for the following functions. $$y=\frac{\sin x}{1+\cos x}$$

Short answer

Question: Determine the derivative of the function \(y=\frac{\sin x}{1+\cos x}\) with respect to x. Answer: The derivative of the function \(y=\frac{\sin x}{1+\cos x}\) with respect to x is \(\frac{dy}{dx}=\frac{1}{1+\cos x}\).

Step-by-step solution

Find \(u'(x)\) and \(v'(x)\)

First, we need to find the derivatives of u(x) and v(x) with respect to x. Using the basic differentiation rules: - \(u(x) = \sin x \Rightarrow u'(x) = \cos x\) - \(v(x) = 1 + \cos x \Rightarrow v'(x) = -\sin x\) We will use these in the next step to apply the quotient rule.

Apply the Quotient Rule

Using the quotient rule from the analysis, we substitute u(x), u'(x), v(x), and v'(x) into the formula: $$\frac{dy}{dx}=\frac{(\cos x)(1+\cos x)-(\sin x)(-\sin x)}{[(1+\cos x)]^2}$$

Simplify the numerator and denominator

Now we expand the numerator and simplify the resulting expression: $$\frac{dy}{dx}=\frac{\cos x+\cos^2 x+\sin^2 x}{(1+\cos x)^2}$$ Recall that \(\sin^2 x + \cos^2 x = 1\), therefore: $$\frac{dy}{dx}=\frac{1+\cos x}{(1+\cos x)^2}$$

Final simplification

Finally, cancel out the (1 + cos x) term in both numerator and denominator: $$\frac{dy}{dx}=\frac{1}{1+\cos x}$$ So the derivative of the given function is: $$\frac{dy}{dx}=\frac{1}{1+\cos x}$$

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which one quantity changes with respect to another. This process is essential in understanding the dynamic behavior of functions. When we differentiate a function, we obtain its derivative, which gives us the slope of the tangent line at any point on the graph of the function.
For example, in calculus, differentiating a trigonometric function like \(\sin x\) gives us \(\cos x\). This tells us that at any point on the sine curve, the rate of change or the slope of the tangent is given by the cosine at that point.

In the exercise where we find \(dy/dx\) for \(y = \frac{\sin x}{1 + \cos x}\), we first differentiate \(\sin x\) and \(1 + \cos x\) separately. Using basic differentiation rules, we find their derivatives: \(\cos x\) for \(\sin x\) and \(-\sin x\) for \(1 + \cos x\).
Understanding differentiation is key to solving many mathematical problems and provides insight into the function's behavior, making it easier to analyze and predict outcomes.

Quotient Rule

The quotient rule is a technique in calculus used to differentiate functions that are expressed as a quotient or ratio of two functions. This rule is particularly useful when dealing with divisions of functions where both the numerator and the denominator are not constants.
The formula for the quotient rule is:\[ \frac{u(x)}{v(x)} \rightarrow \frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \]
In this exercise, the given function \(y = \frac{\sin x}{1+\cos x}\) is a classic example of a problem where the quotient rule applies. Here, \(u(x) = \sin x\) and \(v(x) = 1+\cos x\). We first find the derivatives of \(u\) and \(v\), which are \(\cos x\) and \(-\sin x\), respectively.

Next, we apply the quotient rule by substituting these values into the formula. This involves multiplying and subtracting the terms carefully: \(\cos x(1+\cos x) - \sin x(-\sin x)\), then divide by \((1+\cos x)^2\).
Finally, simplifying the expression guides us towards the solution, unveiling how derivatives can be neatly obtained even in complex functions through systematic application of indelible principles like the quotient rule.

Trigonometric Functions

Trigonometric functions, such as \(\sin x\) and \(\cos x\), form the backbone of various applications in mathematics, physics, and engineering. These functions relate angles of triangles to ratios of side lengths and are related to circular motion.
In our exercise, the function \(y = \frac{\sin x}{1 + \cos x}\) incorporates both sine and cosine. Differentiating functions that involve trigonometric components often requires combining trigonometric identities or rules, such as recognizing that \(\sin^2 x + \cos^2 x = 1\).
This identity was crucial in simplifying the derivative of our given function, allowing us to put it into a more understandable form.

Additionally, derivatives of trigonometric functions allow for analyzing oscillatory behavior and wave patterns, such as sound waves or light waves.\
The knowledge of these functions and how to differentiate them can greatly aid in finding solutions to problems that describe periodic phenomena. Throughout calculus, understanding trigonometric functions and their properties typically leads to deeper insights into complex systems and their changes over time.