Question

If \(y = 600 e^{-7x} + 500 e^{7x}\), then show that \(\frac{d^2y}{dx^2} = 49y\).

Hint:

For functions of the form \(y = A e^{kx} + B e^{-kx}\), the second derivative will always be \(\frac{d^2y}{dx^2} = k^2 y\). Recognizing this pattern can save time in multiple-choice questions or during verification steps.

Answer 1

Step 1: Understanding the Concept:
The problem requires finding the second derivative of the given function \(y\) with respect to \(x\) and then showing that the resulting expression is equal to 49 times the original function \(y\). This involves applying the rules of differentiation for exponential functions.
Step 2: Key Formula or Approach:
The key formula for differentiation is the chain rule applied to exponential functions: \[ \frac{d}{dx}(e^{kx}) = k \cdot e^{kx} \] We will apply this rule twice to find the second derivative.
Step 3: Detailed Explanation or Calculation:
Given the function: \[ y = 600 e^{-7x} + 500 e^{7x} \] First, find the first derivative, \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{d}{dx}(600 e^{-7x} + 500 e^{7x}) \] \[ \frac{dy}{dx} = 600 \cdot (-7)e^{-7x} + 500 \cdot (7)e^{7x} \] \[ \frac{dy}{dx} = -4200 e^{-7x} + 3500 e^{7x} \] Next, find the second derivative, \(\frac{d^2y}{dx^2}\), by differentiating \(\frac{dy}{dx}\): \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-4200 e^{-7x} + 3500 e^{7x}) \] \[ \frac{d^2y}{dx^2} = -4200 \cdot (-7)e^{-7x} + 3500 \cdot (7)e^{7x} \] \[ \frac{d^2y}{dx^2} = 29400 e^{-7x} + 24500 e^{7x} \] Now, let's factor out 49 from this expression: \[ \frac{d^2y}{dx^2} = 49 (600 e^{-7x} + 500 e^{7x}) \] We can see that the expression in the parenthesis is the original function \(y\).
Therefore, \[ \frac{d^2y}{dx^2} = 49y \] Step 4: Final Answer:
By differentiating the function \(y\) twice, we have shown that \(\frac{d^2y}{dx^2}\) is equal to \(49y\), thus verifying the given relation.