Problem 39 - Thomas Calculus

Question

In Exercises \(37-40,\) find \(d y / d x\). $$ e^{2 x}=\sin (x+3 y) $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{2e^{2x} - \cos(x + 3y)}{3\cos(x + 3y)} \)

Step 1: Differentiate Both Sides With Respect to x

To find \( \frac{dy}{dx} \), start by differentiating both sides of the equation \( e^{2x} = \sin(x + 3y) \) with respect to \( x \). For the left side, use the chain rule: the derivative of \(e^{2x}\) is \(2e^{2x}\).

Step 2: Differentiate the Right Side Using Chain Rule and Implicit Differentiation

On the right side, apply the chain rule and implicit differentiation: the derivative of \(\sin(x+3y)\) is \(\cos(x + 3y)\cdot (1 + 3\frac{dy}{dx})\).

Step 3: Set Up the Equation

Equate the derivatives obtained from both sides: \[ 2e^{2x} = \cos(x + 3y)(1 + 3\frac{dy}{dx}) \].

Step 4: Solve for \( \frac{dy}{dx} \)

Rearrange the equation to solve for \( \frac{dy}{dx} \): \[ 3\cos(x + 3y)\frac{dy}{dx} = 2e^{2x} - \cos(x + 3y) \]. Divide both sides by \( 3\cos(x + 3y)\) to isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{2e^{2x} - \cos(x + 3y)}{3\cos(x + 3y)} \].

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus that helps us differentiate composite functions. When you have a function composed of other functions, like in our problem with expressions like \(e^{2x}\) and \(\sin(x + 3y)\), you'll need the chain rule to find their derivatives.

• For \(e^{2x}\), the outer function is the exponential \(e^u\) where \(u = 2x\), and the inner function is \(2x\). The chain rule tells us to differentiate the outer function with respect to the inner one: \(d(e^u)/du = e^u\), and then multiply it by the derivative of the inner function \(d(2x)/dx = 2\). This gives us \(2e^{2x}\).
• The same principle applies for \(\sin(x + 3y)\). With \(x + 3y\) as the inner function, the chain rule first applies to the outer trigonometric function. The derivative of \(\sin(u)\) is \(\cos(u)\), with \(u = x + 3y\). Then, we multiply by the derivative of the inner part, derived using implicit differentiation for \(y\), leading to \(\cos(x + 3y)(1 + 3\frac{dy}{dx})\).

Learning to apply the chain rule allows you to handle complex derivatives systematically, breaking them down step-by-step.

Derivative

Derivatives represent how a function changes as its input changes. They are essential in understanding the behavior of mathematical functions and are widely applicable in various fields such as physics, engineering, and economics.
In our example, when we take the derivative of both sides of the equation \(e^{2x} = \sin(x + 3y)\), we are essentially looking at how each part changes with respect to \(x\).

• The derivative of \(e^{2x}\) with respect to \(x\) is \(2e^{2x}\), as derived using the chain rule.
• For \(\sin(x + 3y)\), its derivative involves using the chain rule and the principle of implicit differentiation, resulting in \(\cos(x + 3y)(1 + 3\frac{dy}{dx})\).

The process of differentiation helps transform an equation into one that describes a rate of change, providing deeper insights into dynamic systems.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent are core concepts in both geometry and calculus. These functions are crucial when you deal with angles and periodic phenomena.
In trigonometric differentiation, the derivatives involve identifying and applying known derivatives of these functions:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• Through the chain rule, if you have something like \(\sin(x + 3y)\), you treat \(x + 3y\) as a single variable and differentiate accordingly, leading to \(\cos(x + 3y)\cdot (1 + 3\frac{dy}{dx})\).

Understanding trigonometric functions and their derivatives allows you to analyze wave patterns and circular motions comprehensively. Differentiating these functions requires careful attention to both their composition and the application of calculus rules like the chain rule. This skill is fundamental when dealing with any kind of oscillation in both theoretical work and real-world applications.