Answer
The derivative of the function \( y = (2x^2 - 8x)e^{-8x} \) is:
\[
y' = (-16x^2 + 68x - 8)e^{-8x}
\]
\end{Output}
Solution
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 \( x \).
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