Question

Find \(y^{\prime \prime}\) for the following functions. $$y=e^{x} \sin x$$

Short answer

Answer: The second derivative of the function is \(y^{\prime \prime} = 2e^x \cos x - 2e^x \sin x\).

Step-by-step solution

Find the first derivative using the product rule

First, recall the product rule for differentiation: \((fg)^\prime = f^\prime g + fg^\prime\). We are given the function \(y = e^x \sin x\), which we can view as the product of two functions, \(f(x) = e^x\) and \(g(x) = \sin x\). Differentiate \(f(x)\) with respect to \(x\): $$f^\prime (x) = \frac{d}{dx} (e^x) = e^x$$ Differentiate \(g(x)\) with respect to \(x\): $$g^\prime (x) = \frac{d}{dx} (\sin x) = \cos x$$ Now apply the product rule formula: $$y^\prime = f^\prime g + fg^\prime = (e^x)(\sin x) + (e^x)(\cos x) = e^x \sin x + e^x \cos x$$

Find the second derivative using the product rule

Now that we have the first derivative \(y^\prime = e^x \sin x + e^x \cos x\), we need to find the second derivative \(y^{\prime \prime}\). We will use the product rule again. Notice that \(y^\prime\) can be viewed as the sum of two products of functions, \(h(x) = e^x\) and \(i(x) = \sin x\), and \(j(x)=e^x\) and \(k(x)=\cos x\), as follows: $$y^\prime = h(x)i(x) + j(x)k(x)$$ Recall that we have already found the derivatives individually for \(h(x)=e^x\) and \(j(x)=e^x\) (\(h^\prime(x)=e^x\) and \(j^\prime(x)=e^x\)). Differentiate \(i(x)\) and \(k(x)\) with respect to \(x\): $$i^\prime (x) = \frac{d}{dx} (\sin x) = \cos x$$ $$k^\prime (x) = \frac{d}{dx} (\cos x) = -\sin x$$ Now we will apply the product rule to \(h(x)i(x)\) and \(j(x)k(x)\) individually. The second derivative will be the sum of the derivative of these two products: $$y^{\prime \prime} = (h^\prime i + hi^\prime) + (j^\prime k + jk^\prime) = (e^x)(\cos x) + (e^x)(\cos x) + (e^x)(-\sin x) + (e^x)(-\sin x)$$ $$y^{\prime \prime} = 2e^x \cos x - 2e^x \sin x$$ So the second derivative of the given function, \(y = e^x \sin x\), is: $$y^{\prime \prime} = 2e^x \cos x - 2e^x \sin x$$

Key concepts

Product Rule

When you are tasked with differentiating a function that is the product of two or more smaller functions, the product rule comes into play. The product rule states that the derivative of a product of two functions, say \(f(x)\) and \(g(x)\), is given by \((fg)' = f'g + fg'\). This can be visualized as each function taking turns being differentiated while the other remains unchanged.

- Think of it as distributing the derivative operation across each component.- It becomes essential when dealing with functions multiplied together, as direct differentiation is not possible.- Applying this rule is crucial in finding derivatives of more complex functions such as in the transcendental example \(y = e^x \sin x\). The key is to break down the expression into manageable parts, then individually find their derivatives. Once obtained, you can reconstruct the derivative of the entire product by applying the product rule formula.

Differentiation

Differentiation is a cornerstone operation in calculus that helps in finding the rate at which a value changes with respect to a variable. This process is crucial for comprehending how functions behave as their inputs vary, effectively revealing their slopes at any given point.- Differentiation involves applying specific rules to determine derivatives. These derivatives express the function's instantaneous rate of change.- For common functions, rules like the power rule, quotient rule, and product rule simplify finding their derivatives.- Functions such as \(e^x\) and \(\sin x\) have well-known derivatives \(e^x\) and \(\cos x\) respectively.In the exercise given, differentiating starts with using the product rule for the composite function \(y = e^x \sin x\). By finding first and second derivatives, we comprehend how the function transforms and curves in space. Each step builds on the previous one, showcasing the cumulative nature of calculus.

Transcendental Functions

Transcendental functions go beyond algebraic expressions involving polynomials, sines, cosines, logarithms, and exponentials. They often appear in mathematical, engineering, and scientific calculations due to their ability to model real-world phenomena.- Common examples include functions like \(e^x\) and \(\sin x\), present in varying fields from physics to engineering.- They do not follow algebraic solutions and often require calculus tools like differentiation to analyze.- When differentiating transcendental functions, recognizing their unique derivatives is essential for correct application.In our exercise, you encountered \(y = e^x \sin x\), a function that combines both exponentials and trigonometric terms. To find its derivative accurately, using differentiation rules pertinent to transcendental functions, like the product rule, is vital. Understanding these functions' properties and behaviors supports comprehensive calculus problem-solving.