y=(3x−2).e−2x𝑦=(3𝑥−2).𝑒−2𝑥   Giá trị của  y′′(1)𝑦″(1)  là:Select one:a.y′′(1)=−8e2𝑦"(1)=−8𝑒2b.y′′(1)=−7e2𝑦"(1)=−7𝑒2c.y′′(1)=−8e−2𝑦"(1)=−8𝑒−2d.y′′(1)=8e2

1. Break Down the Problem

We need to find the second derivative $y''(1)$ for the function given by:

$y = (3x - 2)e^{-2x}$

Once we find $y''(1)$, we will compare it to the options provided to select the correct answer.

2. Relevant Concepts

To find $y''(x)$, we will need to:

• Find the first derivative $y'$ using the product rule.
• Then, find the second derivative $y''$ also using the product rule and possibly the chain rule.
• Evaluate $y''(1)$.

3. Analysis and Detail

Finding the first derivative $y'$:

Using the product rule, where $u = 3x - 2$ and $v = e^{-2x}$:

$y' = u'v + uv'$

Calculating $u'$ and $v'$:

• $u' = 3$
• $v = e^{-2x}$
• $v' = -2e^{-2x}$

Now substituting into the product rule:

$y' = (3)(e^{-2x}) + (3x - 2)(-2e^{-2x})$

Simplifying $y'$:

$y' = 3e^{-2x} - 2(3x - 2)e^{-2x} = (3 - 6x + 4)e^{-2x} = (-6x + 7)e^{-2x}$

Finding the second derivative $y''$:

Now differentiate $y'$ again using the product rule:

Let $u = -6x + 7$ and $v = e^{-2x}$:

$y'' = u'v + uv'$

Calculating $u'$ and $v'$:

• $u' = -6$
• $v' = -2e^{-2x}$

Substituting:

$y'' = (-6)(e^{-2x}) + (-6x + 7)(-2e^{-2x})$

Simplifying:

$y'' = -6e^{-2x} + 12(3x - \frac{7}{2})e^{-2x}$

Putting it all together:

$y'' = (-6 + 12(3x - 7/2))e^{-2x}$

4. Verify and Summarize

Now evaluate $y''(1)$:

$y''(1) = (-6 + 12(3 \cdot 1 - 7/2))e^{-2 \cdot 1}$ $= (-6 + 12(3 - 3.5))e^{-2} = (-6 + 12(-0.5))e^{-2} = (-6 - 6)e^{-2} = -12e^{-2}$

Final Answer

After thoroughly calculating $y''(1)$ we have:

$y''(1) = -12e^{-2}$

None of the options exactly match the computed result $-12e^{-2}$. Thus, based on the available options, there may have been an error or mismatch in the options presented. However, the calculated value is provided for your reference.