Find the derivative of the given function. \[ y=\left(2 x^{2}-8 x\right) e^{-8 x} \]

Alright, I need to find the derivative of the function \( y = (2x^2 - 8x)e^{-8x} \). Let's break this down step by step to ensure I cover all possible angles and methods. First, I recognize that this is a product of two functions: \( u(x) = 2x^2 - 8x \) and \( v(x) = e^{-8x} \). To find the derivative of their product, I'll use the product rule, which states that if \( y = u(x)v(x) \), then \( y' = u'(x)v(x) + u(x)v'(x) \). So, let's find the derivatives of \( u(x) \) and \( v(x) \) individually. 1. **Finding \( u'(x) \):** \( u(x) = 2x^2 - 8x \) - The derivative of \( 2x^2 \) with respect to \( x \) is \( 4x \). - The derivative of \( -8x \) with respect to \( x \) is \( -8 \). - Therefore, \( u'(x) = 4x - 8 \). 2. **Finding \( v'(x) \):** \( v(x) = e^{-8x} \) - The derivative of \( e^{kx} \) with respect to \( x \) is \( ke^{kx} \), where \( k \) is a constant. - Here, \( k = -8 \), so \( v'(x) = -8e^{-8x} \). Now, applying the product rule: \[ y' = u'(x)v(x) + u(x)v'(x) \] Substituting the derivatives we found: \[ y' = (4x - 8)e^{-8x} + (2x^2 - 8x)(-8e^{-8x}) \] Let's simplify this expression step by step. First, distribute the terms: \[ y' = (4x - 8)e^{-8x} - 8(2x^2 - 8x)e^{-8x} \] \[ y' = (4x - 8)e^{-8x} - 16x^2e^{-8x} + 64xe^{-8x} \] Now, combine like terms: \[ y' = (4x - 8 - 16x^2 + 64x)e^{-8x} \] \[ y' = (-16x^2 + 68x - 8)e^{-8x} \] To make it more readable, I can factor out a common factor of \( -8 \) from the polynomial: \[ y' = -8(2x^2 - 8.5x + 1)e^{-8x} \] Alternatively, if I prefer to keep it in the expanded form: \[ y' = (-16x^2 + 68x - 8)e^{-8x} \] Both forms are correct, but the factored form might be more useful depending on the context or further simplification needed. **Double-Checking:** To ensure there are no mistakes, I'll verify the derivative by plugging in a value for \( x \) and checking if the derivative makes sense. Let's choose \( x = 1 \): \[ u(1) = 2(1)^2 - 8(1) = 2 - 8 = -6 \] \[ v(1) = e^{-8(1)} = e^{-8} \approx 0.000335 \] \[ u'(1) = 4(1) - 8 = 4 - 8 = -4 \] \[ v'(1) = -8e^{-8} \approx -8 \times 0.000335 \approx -0.00268 \] \[ y'(1) = (-4)(0.000335) + (-6)(-0.00268) \approx -0.00134 + 0.01608 \approx 0.01474 \] Calculating the derivative using the expanded form: \[ y'(1) = (-16(1)^2 + 68(1) - 8)e^{-8} = (-16 + 68 - 8)e^{-8} = 44e^{-8} \approx 44 \times 0.000335 \approx 0.01474 \] The results match, confirming that the derivative is correctly calculated. **Alternative Approach:** Another way to approach this problem is by using logarithmic differentiation, which can be particularly useful for products and quotients of functions. Starting with: \[ y = (2x^2 - 8x)e^{-8x} \] Take the natural logarithm of both sides: \[ \ln(y) = \ln(2x^2 - 8x) + \ln(e^{-8x}) = \ln(2x^2 - 8x) - 8x \] Differentiate both sides with respect to \( x \): \[ \frac{1}{y} \cdot y' = \frac{4x - 8}{2x^2 - 8x} - 8 \] Multiply both sides by \( y \): \[ y' = y \left( \frac{4x - 8}{2x^2 - 8x} - 8 \right) \] Substitute \( y = (2x^2 - 8x)e^{-8x} \): \[ y' = (2x^2 - 8x)e^{-8x} \left( \frac{4x - 8}{2x^2 - 8x} - 8 \right) \] Simplify the expression inside the parentheses: \[ \frac{4x - 8}{2x^2 - 8x} = \frac{4(x - 2)}{2x(x - 4)} = \frac{2(x - 2)}{x(x - 4)} \] So, \[ y' = (2x^2 - 8x)e^{-8x} \left( \frac{2(x - 2)}{x(x - 4)} - 8 \right) \] This approach leads to the same derivative as the product rule method, confirming the result. **Conclusion:** After applying both the product rule and logarithmic differentiation, the derivative of the function \( y = (2x^2 - 8x)e^{-8x} \) is: \[ y' = (-16x^2 + 68x - 8)e^{-8x} \] This derivative accurately represents the rate of change of the original function with respect to \( x \).