Problem 16 - A Textbook of Differential Calculus

Question

The functions \(u=e^{x} \sin x, v=e^{x} \cos x\) satisfy the equation (a) \(v \frac{d u}{d x}-u \frac{d v}{d x}=u^{2}+v^{2}\) (b) \(\frac{d^{2} u}{d x^{2}}=2 v\) (c) \(\frac{d^{2} v}{d x^{2}}=-2 u\) (d) All of these

Step-by-Step Solution

Answer

The answer is (d) All of these.

Step 1: Compute the First Derivative of u

Given the function \( u = e^x \sin x \). Apply the product rule to differentiate it. The derivative is: \[ \frac{du}{dx} = \frac{d}{dx}(e^x \sin x) = e^x \sin x + e^x \cos x \] The result is \( \frac{du}{dx} = e^x (\sin x + \cos x) \).

Step 2: Compute the First Derivative of v

Given the function \( v = e^x \cos x \). Apply the product rule to find the derivative. The derivative is: \[ \frac{dv}{dx} = \frac{d}{dx}(e^x \cos x) = e^x \cos x - e^x \sin x \] The result is \( \frac{dv}{dx} = e^x (\cos x - \sin x) \).

Step 3: Substituting into Given Equation (a)

We need to verify equation (a), which is \( v \frac{d u}{d x}-u \frac{d v}{d x}=u^{2}+v^{2} \). Using previously computed derivatives and original functions: \[ u = e^x \sin x \quad v = e^x \cos x \] Substitute: \[ v \frac{du}{dx} - u \frac{dv}{dx} = e^x \cos x \cdot e^x (\sin x + \cos x) - e^x \sin x \cdot e^x (\cos x - \sin x) \] Simplify to: \[ = (e^{2x})(\cos x \sin x + \cos^2 x) - (e^{2x})(\sin x \cos x - \sin^2 x) \] \[ = e^{2x} (\cos^2 x + \sin^2 x) = e^{2x} \] Also \[ u^2 + v^2 = (e^x \sin x)^2 + (e^x \cos x)^2 = e^{2x} (\sin^2 x + \cos^2 x) = e^{2x} \], hence the equation is satisfied.

Step 4: Compute the Second Derivative of u

We now compute \( \frac{d^2 u}{dx^2} \). Differentiate \( \frac{du}{dx} = e^x (\sin x + \cos x) \) again: \[ \frac{d^2 u}{dx^2} = \frac{d}{dx}(e^x \sin x + e^x \cos x) = e^x \sin x + e^x \cos x + e^x \cos x - e^x \sin x = 2e^x \cos x = 2v \] Therefore, \( \frac{d^2 u}{dx^2} = 2v \), which satisfies equation (b).

Step 5: Compute the Second Derivative of v

Differentiate \( \frac{dv}{dx} = e^x (\cos x - \sin x) \) for \( \frac{d^2 v}{dx^2} \): \[ \frac{d^2 v}{dx^2} = \frac{d}{dx}(e^x \cos x - e^x \sin x) = e^x \cos x - e^x \sin x - (e^x \sin x + e^x \cos x) = -2e^x \sin x = -2u \] Hence, \( \frac{d^2 v}{dx^2} = -2u \), which satisfies equation (c).

Step 6: Conclusion: Verify All Options

We have shown that all three conditions (a), (b), and (c) are satisfied. Therefore, the correct answer is (d) All of these.

Key Concepts

Product Rule

The product rule is a fundamental principle used in differential calculus to find derivatives of functions that are products of two simpler functions. If we have a function defined as the product of two functions, say \( f(x) = u(x) \cdot v(x) \), then the product rule states that the derivative of \( f(x) \), denoted as \( f'(x) \), is:

• \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)
• Here, \( u'(x) \) and \( v'(x) \) are the derivatives of \( u(x) \) and \( v(x) \) respectively.

This rule helps us differentiate products like \( e^x \sin x \) and \( e^x \cos x \), as seen in the problem.
To apply the product rule, always:

• Differentiate the first function and multiply it by the second function.
• Then add the product of the first function and the derivative of the second function.

This systematic approach makes it easy to handle products in differentiation.

First Derivative

The first derivative of a function represents the rate of change of the function with respect to its variable. In simple terms, it tells us how the function's value changes as the input changes. This concept is crucial for understanding how functions behave.
In the context of this exercise, we computed the first derivatives for two functions, \( u = e^x \sin x \) and \( v = e^x \cos x \), using the product rule:

• First derivative of \( u \): \( \frac{du}{dx} = e^x (\sin x + \cos x) \)
• First derivative of \( v \): \( \frac{dv}{dx} = e^x (\cos x - \sin x) \)

These derivatives are used to analyze the original equation given in the exercise.
Knowing how to find the first derivative allows us to further investigate a function’s behavior, like its increasing or decreasing nature, and prepare us for more complex analyses like finding second derivatives.

Second Derivative

The second derivative of a function provides information about the concavity of the function and its inflection points. It is the derivative of the first derivative, essentially analyzing how the rate of change itself is changing.
In this exercise, we find the second derivatives of \( u \) and \( v \):

• \( \frac{d^2 u}{dx^2} = 2v \)
• \( \frac{d^2 v}{dx^2} = -2u \)

These results are critical for verifying equations (b) and (c) from the exercise.
The second derivative can tell us whether a function is concave up or concave down:

• If \( \frac{d^2 f}{dx^2} > 0 \), the function is concave up.
• If \( \frac{d^2 f}{dx^2} < 0 \), the function is concave down.

By understanding the second derivative, we gain insight not just into the function itself but also into the behavior of its rate of change.

Trigonometric Functions

Trigonometric functions like \( \sin(x) \) and \( \cos(x) \) appear frequently in calculus problems. They are periodic functions that model cyclical patterns such as waves. Understanding their properties and how they behave under differentiation is essential in calculus.
Important derivatives of trigonometric functions include:

• \( \frac{d}{dx} \sin x = \cos x \)
• \( \frac{d}{dx} \cos x = -\sin x \)

These derivatives play a significant role in solving the differential calculus exercises as seen where:

• The function \( u = e^x \sin x \) uses \( \sin x \) which when differentiated gives \( \cos x \).
• The function \( v = e^x \cos x \) uses \( \cos x \) which when differentiated gives \(-\sin x \).

Grasping these concepts helps in differentiating products involving exponential and trigonometric components, as observed in the given problem.