Question

In Exercises \(1-8,\) use the given substitution and the Chain Rule to find \(d y / d x\) $$y=\left(\frac{\sin x}{1+\cos x}\right)^{2}, \quad u=\frac{\sin x}{1+\cos x}$$

Short answer

The derivative of the given function is \( \frac {dy} {dx} = \frac{2 \sin x}{(1+ \cos x)^2} \)

Step-by-step solution

Find \(dy / du\)

First, replace \( x \) in the function \( y \) with \( u \) which yields \(y = u^2 \). The derivative of the function with respect to \( u \) (written as \( \frac{dy}{du} \)) is given by using power rule of derivatives:\[ \frac{dy}{du} = 2u\]

Find \(du / dx\)

Now, differentiate \( u = \frac{\sin x}{1+ \cos x} \) with respect to \( x \) yields: \[ \frac{du}{dx} = \frac{(1+\cos x) \cdot \cos x - \sin x \cdot (-\sin x)}{(1+ \cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1+ \cos x)^2} = \frac{1 + \cos x}{(1+ \cos x)^2} = \frac{1}{1+ \cos x}\]

Apply the Chain Rule

The Chain Rule states that \( \frac {dy} {dx} = \frac {dy} {du} \frac {du} {dx} \). Substitute \( \frac {dy} {du} \) from step 1 and \( \frac {du} {dx} \) from step 2:\[ \frac {dy} {dx} = 2u \cdot \frac{1}{1+ \cos x} = \frac{2 \sin x}{(1+ \cos x)^2}\]

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used when variables cannot be easily separated for direct differentiation, such as in functions where x and y are mixed together. In our exercise, we use implicit differentiation indirectly. While the given function appears to be in explicit form, our substitution introduces a new variable, u, which necessitates the use of implicit differentiation methods.

During this process, we treat y as an implicit function of both x and u, since y depends on u, which in turn depends on x. By calculating the derivatives \( \frac{dy}{du} \) and \( \frac{du}{dx} \) separately, we're able to implicitly find \( \frac{dy}{dx} \) by combining these through the chain rule. This approach simplifies the differentiation process by breaking it into more manageable steps.

Trigonometric Functions

Trigonometric functions are fundamental in calculus, as they describe the relationship between the angles and sides of triangles. They are also periodic and have various properties that can be utilized in differentiation and integration. In our exercise, the functions sine and cosine appear both in the original function for y and the substitution for u.

It's important to recognize that trigonometric functions have derivatives that are cyclical. Specifically, the derivative of the sine function (sin x) is the cosine function (cos x), and the derivative of the cosine function (cos x) is minus the sine function (-sin x). Understanding these derivatives is essential when using the chain rule and performing implicit differentiation in calculus.

Substitution Method in Calculus

The substitution method in calculus is a technique used to simplify the process of finding derivatives and integrals. Essentially, it involves replacing a complicated expression with a single variable, making the calculation easier to perform.

In the exercise presented, we have made the substitution \( u = \frac{\sin x}{1+ \cos x} \), which transforms the original complicated function into a simple function of u, \( y = u^2 \). This allows us to apply the power rule directly to find \( \frac{dy}{du} \) before back-substituting to find \( \frac{dy}{dx} \) using the chain rule. Substitution not only simplifies calculations but also provides a clear example of how different parts of calculus are interconnected.

Power Rule for Derivatives

The power rule is one of the most fundamental rules of differentiation. It states that if you have a function in the form of \( y = u^n \), where n is a real number, the derivative of y with respect to u is \( \frac{dy}{du} = n\cdot u^{n-1} \).

In our exercise, once we performed the substitution to make \( y = u^2 \), we applied the power rule to find \( \frac{dy}{du} = 2u \) (since \( u^2 \) is in the form of \( u^n \) where \( n = 2 \)). This rule is crucial because it can be applied quickly and to many functions with exponential forms. Thus, understanding and applying the power rule is a key skill in calculus that significantly eases the process of differentiation.