Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(y=\sin \left(x^{2}\right)+2 x^{3}\), find \(d x / d y\).

Step-by-Step Solution

Answer

\( \frac{dx}{dy} = \frac{1}{2x \cos(x^2) + 6x^2} \).

Step 1: Identify the Given Function

Given function is \( y = \sin(x^2) + 2x^3 \). We need to find \( \frac{dx}{dy} \), the reciprocal of the derivative \( \frac{dy}{dx} \).

Step 2: Differentiate with Respect to x

Find \( \frac{dy}{dx} \) by differentiating the given function. For \( \sin(x^2) \), use the chain rule: the derivative is \( 2x \cos(x^2) \). For \( 2x^3 \), the derivative is \( 6x^2 \). Thus, \[ \frac{dy}{dx} = 2x \cos(x^2) + 6x^2 \].

Step 3: Find the Reciprocal

We need \( \frac{dx}{dy} \), which is the reciprocal of \( \frac{dy}{dx} \). Hence, \[ \frac{dx}{dy} = \frac{1}{2x \cos(x^2) + 6x^2} \].

Key Concepts

Chain Rule

The chain rule is a fundamental technique used in calculus to differentiate composite functions. It helps us break down the process of finding derivatives when a function is nested inside another function. For example, consider the function \( y = \sin(x^2) + 2x^3 \) from the original exercise. Here, \( \sin(x^2) \) is a composite function, as it involves the inner function \( x^2 \) nested within the outer function \( \sin(u) \) where \( u = x^2 \).
To differentiate \( \sin(x^2) \) using the chain rule, first differentiate the outer function \( \sin(u) \) with respect to \( u \), which gives us \( \cos(u) \). Next, differentiate the inner function \( x^2 \) with respect to \( x \), which gives \( 2x \). Multiply these two results to obtain the derivative of the composite function \( 2x \cos(x^2) \).
Thus, the chain rule allows us to efficiently find the derivative of complex layered functions using simpler, separate derivatives.

Reciprocal Derivative

The concept of a reciprocal derivative is crucial when you need to calculate the inverse of a given derivative. From the exercise, we were tasked with finding \( \frac{dx}{dy} \) rather than the regular derivative. This involves computing the reciprocal of \( \frac{dy}{dx} \).
After differentiating the given function \( y = \sin(x^2) + 2x^3 \) to find \( \frac{dy}{dx} \), which resulted in \( 2x \cos(x^2) + 6x^2 \), the next step is to take its reciprocal to get \( \frac{dx}{dy} \). Thus, \( \frac{dx}{dy} = \frac{1}{2x \cos(x^2) + 6x^2} \).
This reciprocal relationship reflects how changes in \( y \) relate to changes in \( x \), emphasizing the inverse nature of these derivatives.

Derivatives

Derivatives form the backbone of calculus, representing the rate of change of a function. In the exercise, the derivative \( \frac{dy}{dx} \) tells us how the function \( y = \sin(x^2) + 2x^3 \) changes with respect to \( x \).
Taking the derivative of a function requires applying rules like the power rule, product rule, and chain rule. Each rule serves a specific type of function or combination of functions to yield the rate at which they change.
In this particular scenario, differentiating \( 2x^3 \) uses the power rule, giving \( 6x^2 \), and differentiating \( \sin(x^2) \) uses the chain rule to get \( 2x \cos(x^2) \).
Thus, derivatives allow us to understand and calculate the momentum of a function across various points.

Calculus

Calculus is the mathematical study of continuous change, essentially broken down into two main branches: differential calculus and integral calculus. In this exercise, we are delving into differential calculus, focusing on derivatives and differentiating functions.
Differential calculus provides tools for understanding how a function progresses at any given point, offering insights into slopes, tangents, and rates of change. It's particularly useful in fields ranging from physics and engineering to economics and biology.
By mastering concepts like derivatives, chain rule, and reciprocal of derivatives, you can tackle a wide array of problems involving dynamic behaviors or optimization.
The beauty of calculus lies in its ability to break down complex phenomena into understandable units, aiding in the development of solutions to real-world challenges.