Question

Find \(y^{\prime \prime}\) for the following functions. $$y=\frac{1}{2} e^{x} \cos x$$

Short answer

Answer: The second derivative of the function is \(y'' = - e^{x} \sin x\).

Step-by-step solution

Find the first derivative

To find \(y'\), apply the product rule for differentiation of the function \(y=\frac{1}{2} e^{x} \cos x\), which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. In this case, let \(u=\frac{1}{2} e^{x}\) and \(v=\cos x\). First, find the derivatives of \(u\) and \(v\): $$u' = \frac{d}{dx}\left(\frac{1}{2} e^{x}\right) = \frac{1}{2} e^{x}$$ $$v' = \frac{d}{dx}(\cos x) = -\sin x$$ Now, apply the product rule: $$y' = u'v + uv' = \left(\frac{1}{2} e^{x}\right)(\cos x) + \left(\frac{1}{2} e^{x}\right)(-\sin x)$$

Simplify the first derivative

Combine the terms in the first derivative: $$y' = \frac{1}{2} e^{x} \cos x - \frac{1}{2} e^{x} \sin x$$

Find the second derivative

To find \(y''\), apply the product rule for differentiation just like in Step 1. Let \(u=\frac{1}{2} e^{x}\) and \(v=\cos x - \sin x\). We already have the derivatives of \(u\) and \(v\) from Step 1, so we can use them directly to find the second derivative. Apply the product rule: $$y'' = u'v + uv' = \left(\frac{1}{2} e^{x}\right)(\cos x - \sin x) + \left(\frac{1}{2} e^{x}\right)(-\sin x - \cos x)$$

Simplify the second derivative

Combine the terms in the second derivative: $$y'' = \frac{1}{2} e^{x} \cos x - \frac{1}{2} e^{x} \sin x - \frac{1}{2} e^{x} \sin x - \frac{1}{2} e^{x} \cos x$$ Now, simplify the expression by combining like terms: $$y'' = - e^{x} \sin x$$

Key concepts

Product Rule in Calculus

The product rule in calculus is an essential tool for differentiating products of two functions. It's particularly handy when dealing with expressions like the original function you encountered: \( y = \frac{1}{2} e^{x} \cos x \). In this context, the product rule is indispensable.To use the product rule, remember the formula: if you have a product of two functions \( u(x) \) and \( v(x) \), the derivative is given by:

• \( y' = u'v + uv' \)

Here, \( u' \) is the derivative of \( u \) and \( v' \) is the derivative of \( v \). Applying this to our function, we let \( u = \frac{1}{2} e^{x} \) and \( v = \cos x \). We found that \( u' = \frac{1}{2} e^{x} \) and \( v' = -\sin x \).Using the formula, plug in the values to find the first derivative:

• \( y' = \left( \frac{1}{2} e^{x} \right)(\cos x) + \left( \frac{1}{2} e^{x} \right)(-\sin x) \)

The result simplifies to \( y' = \frac{1}{2} e^{x} \cos x - \frac{1}{2} e^{x} \sin x \). Knowing how to apply the product rule correctly is crucial for efficiently handling derivative calculations involving multiple functions.

Exponential Functions

Exponential functions are a crucial concept in calculus and are found in many mathematical expressions. In our example, you dealt with \( e^{x} \), a quintessential exponential function.An exponential function is generally expressed as \( a^{x} \), where \( a \) is a constant, and \( e \, ( \approx 2.71828) \) is the natural base of logarithms. The function grows (or decays) exponentially, meaning it increases rapidly at a rate proportional to its current value.One of the fascinating features of \( e^{x} \) is that its derivative is itself:

• \( \frac{d}{dx}(e^{x}) = e^{x} \)

This property makes \( e^{x} \) exceptionally simple to differentiate and is often a part of composite functions in calculus.In your exercise, \( \frac{1}{2} e^{x} \) was part of the function to differentiate. The simplicity of \( e^{x} \)'s properties helps greatly to quickly compute its derivatives, simplifying the differentiation process.

Trigonometric Functions

Trigonometric functions such as \( \cos x \) and \( \sin x \) frequently appear in calculus problems, especially those involving derivatives and integrals. These functions relate to angles in right triangles and the unit circle.In the function provided, \( \cos x \) served as a component of the product during the differentiation process. Its derivative is a fundamental trigonometric identity:

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

Similarly, you encounter \( \sin x \) while further simplifying our terms:

• \( \frac{d}{dx}(\sin x) = \cos x \)

Understanding these derivatives allows you to manage and simplify functions involving any combination of \( \cos x \) or \( \sin x \).In the example problem, knowledge of these relations enabled the efficient finding of both the first and second derivatives, as combining the derivatives assisted in simplifying the expressions quickly. Recognizing and applying these differences is essential to mastering calculus and helps streamline problem-solving considerably.