Problem 15 - Calculus Early Transcendentals

Question

Use Version I of the Chain Rule to calculate \(\frac{d y}{d x}\). $$y=\tan 5 x^{2}$$

Step-by-Step Solution

Answer

Question: Find the derivative of the function y = tan(5x^2) with respect to x. Answer: The derivative of the function y = tan(5x^2) with respect to x is dy/dx = 10x∙sec^2(5x^2).

Step 1: Identify the outer function and inner function

In this exercise, our given function is y = tan(5x^2). Here, the outer function is tan(u) and the inner function u is 5x^2.

Step 2: Differentiate the outer function with respect to the inner function

The outer function is tan(u). To differentiate this function with respect to u, we have: $$\frac{d(\tan(u))}{du} = \sec^2(u)$$

Step 3: Differentiate the inner function with respect to x

The inner function is u = 5x^2. To differentiate this function with respect to x, we have: $$\frac{d(5x^2)}{dx} = 10x$$

Step 4: Apply the Chain Rule

The chain rule states that: $$\frac{dy}{dx} = \frac{d(\tan(u))}{du} \cdot \frac{du}{dx}$$ Substitute the values we found in Steps 2 and 3 into the equation: $$\frac{dy}{dx} = \sec^2(u) \cdot 10x$$

Step 5: Replace u with the inner function

We need to replace u with the inner function, which is 5x^2: $$\frac{dy}{dx} = \sec^2(5x^2) \cdot 10x$$

Step 6: Simplify the expression

We can rewrite our final expression as: $$\frac{dy}{dx} = 10x\sec^2(5x^2)$$ This is the derivative of y = tan(5x^2) with respect to x.

Key Concepts

Differentiation

Differentiation is a cornerstone concept in calculus, essential for understanding rates of change and the behavior of graphs. It allows us to find the derivative of a function, which represents the function's rate of change at any given point. The derivative can be thought of as the slope of the function's graph at that point. For instance, in a real-world scenario, if we consider the function representing the distance traveled by a car over time, the derivative would give us the car's velocity at any moment.

When we differentiate basic functions, like polynomials, we use simple rules, such as the power rule, which tells us that the derivative of \( x^n \) is \( nx^{n-1} \). However, when functions become more complex, involving products, quotients, or compositions of functions, as it often happens with trigonometric functions, we need to use more advanced techniques such as the product rule, quotient rule, and the chain rule to find their derivatives.

Trigonometric Functions

Trigonometric functions are a class of functions that are fundamental in trigonometry, and they appear widely in calculus as well. These functions, which include sine \( (\sin) \), cosine \( (\cos) \), tangent \( (\tan) \), and their reciprocals, relate the angles of a triangle to the lengths of its sides. They are periodic, and their graphs are waves commonly seen in natural phenomena like sound and light.

Basic Derivatives of Trigonometric Functions

For differentiation purposes, it is important to recall the derivatives of the basic trigonometric functions:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

In our exercise, we specifically deal with the tangent function, whose behavior and rate of change at any given angle can be characterized by using these differentiation principles.

Derivative of Composite Functions

The derivative of composite functions involves taking the derivative of a function that is nested within another. This is where the chain rule becomes an invaluable tool in calculus. The chain rule allows us to differentiate composite functions by breaking down the differentiation process into manageable steps, where we first differentiate the outer function and then multiply it by the derivative of the inner function.

Formula for the Chain Rule

The chain rule can be stated as follows: If you have a composite function \( y = f(g(x)) \), then the derivative of \( y \) with respect to \( x \) is:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \),

where \( f'(g(x)) \) is the derivative of the outer function evaluated at \( g(x) \), and \( g'(x) \) is the derivative of the inner function. In our exercise, the outer function is \( \tan(u) \), and the inner function is \( u = 5x^2 \). By applying the chain rule, we differentiated \( \tan(u) \) with respect to \( u \) and \( 5x^2 \) with respect to \( x \), eventually multiplying the derivatives to find the derivative of the original composite function.