Problem 32 - Single Variable Calculus

Question

Use logarithmic differentiation to find the derivative of the function. $$ y=\frac{\sin ^{2} x}{x^{2} \sqrt{1+\tan x}} $$

Step-by-Step Solution

Answer

The derivative of the function, \(y=\frac{\sin^2 x}{x^2\sqrt{1+\tan x}}\), using logarithmic differentiation is: \[ \frac{dy}{dx}=\frac{\sin x}{x^2 \sqrt{1+\tan x}}\left(2\cos x-\frac{2\sin x}{x}-\frac{\sec^2 x}{2(1+\tan x)}\right). \]

Step 1: Take the natural logarithm of both sides of the equation

Take the natural logarithm of both sides of the equation to simplify the expression: \[ \ln(y)=\ln\left(\frac{\sin^2 x}{x^2\sqrt{1+\tan x}}\right). \]

Step 2: Simplify the expression on the right-hand side using properties of logarithms

Apply the properties of logarithms to simplify the expression on the right-hand side: \[ \ln(y)=\ln(\sin^2 x)-\ln(x^2)-\frac{1}{2}\ln(1+\tan x). \]

Step 3: Differentiate both sides of the equation with respect to x

Differentiate both sides of the equation with respect to x using the chain rule and the fact that \(\frac{d}{dx}[\ln(u)]=\frac{1}{u}\frac{du}{dx}\): \[ \frac{d}{dx}[\ln(y)]=\frac{d}{dx}[\ln(\sin^2 x)]-\frac{d}{dx}[\ln(x^2)]-\frac{1}{2}\frac{d}{dx}[\ln(1+\tan x)]. \] Now let's find the derivatives of each term: \[ \frac{1}{y}\frac{dy}{dx}=2\sin x\cos x\cdot\frac{1}{\sin^2 x}-2x\cdot\frac{1}{x^2}-\frac{1}{2}\cdot\frac{\sec^2 x}{1+\tan x}. \]

Step 4: Solve for the derivative of y with respect to x

Solve for \(\frac{dy}{dx}\) by multiplying both sides of the equation by y: \[ \frac{dy}{dx}=\frac{\sin^2 x}{x^2\sqrt{1+\tan x}}\left(2\cos x\cdot\frac{1}{\sin x}-2\frac{1}{x}-\frac{\sec^2 x}{2(1+\tan x)}\right). \]

Step 5: Simplify the expression for the derivative

Simplify the expression for the derivative: \[ \frac{dy}{dx}=\frac{\sin x}{x^2 \sqrt{1+\tan x}}\left(2\cos x-\frac{2\sin x}{x}-\frac{\sec^2 x}{2(1+\tan x)}\right). \] So the derivative of the function using logarithmic differentiation is: \[ \frac{dy}{dx}=\frac{\sin x}{x^2 \sqrt{1+\tan x}}\left(2\cos x-\frac{2\sin x}{x}-\frac{\sec^2 x}{2(1+\tan x)}\right). \]

Key Concepts

Derivatives of Trigonometric Functions

Understanding the derivatives of trigonometric functions is crucial when dealing with calculus problems involving sine, cosine, tangent, and other trigonometric ratios. These derivatives provide information on how the value of a trigonometric function changes at a certain point.

For example, the derivative of the sine function, sin(x), with respect to x is cos(x). Similarly, the derivative of cos(x) is -sin(x). These are the simplest forms of trigonometric derivatives and are often used as building blocks for more complex functions.

In the given exercise, the function involves sin²(x) and tan(x), which are more complicated than the basic sine and cosine functions. Therefore, we apply principles like the chain rule for differentiation (another concept we'll discuss) to ascertain the derivatives correctly. For the function sin²(x), the derivative is 2sin(x)cos(x), which simplifies to sin(2x), a double angle identity in trigonometry.

Understanding these derivatives and using known trigonometric identities can simplify the process of finding the derivative of a complex trigonometric expression, as shown in the solution steps for our exercise.

Properties of Logarithms

The properties of logarithms play a significant role when we apply logarithmic differentiation. For many students, these properties are sometimes counterintuitive but they are fundamental in simplifying complex expressions.

Some useful properties include:

• The logarithm of a product is the sum of the logarithms of the individual factors: log_b(MN) = log_b(M) + log_b(N).
• The logarithm of a quotient is the difference of the logarithms: log_b(M/N) = log_b(M) - log_b(N).
• The logarithm of a power is the exponent times the logarithm of the base: log_b(M^p) = p log_b(M).

When you take the natural log of both sides of the function in our exercise, you utilize these properties to break down the expression into several components that can be dealt with individually. Especially, converting the exponents and roots into multiplications and divisions allows us to apply the differentiation rules more straightforwardly.

Remembering these properties can help you transform complex expressions into more manageable forms and is a technique often used together with logarithmic differentiation to find derivatives.

Chain Rule for Differentiation

The chain rule for differentiation is an essential tool used to differentiate composite functions. It can be looked at as taking the derivative of the outer function evaluated at the inner function and then multiplying it by the derivative of the inner function.

The general formula for the chain rule can be expressed as if you have a composite function f(g(x)), the derivative, f'(g(x))g'(x), is obtained. In practice, this means if you have a function within another function, you need to take the derivative of the outer function first, while treating the inner function as a constant, and then multiply by the derivative of the inner function.

In the context of the given exercise, when we differentiate ln(y), recognizing y as a function of x, we get the derivative as 1/y dy/dx. We do the same for the other terms that are composed of trigonometric functions inside the natural logarithm.

Mastering the chain rule enables students to tackle a wide range of differentiation problems, especially those involving complicated compositions of functions, including trigonometric and logarithmic functions, as demonstrated in the steps to solve the given exercise.