2.7 Q-2

Question Statement

Find $y_{2}$ for the following expressions:

i. $y = x^{2} e^{-x}$
ii. $y = \ln \left( \frac{2x+3}{3x+2} \right)$

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Background and Explanation

To solve this problem, we need to differentiate the given functions twice with respect to $x$.
The first expression involves a product of two functions, $x^2$ and $e^{-x}$, which requires the product rule for differentiation.
The second expression involves the quotient rule to handle the division inside the logarithm, and the chain rule will also be applied as part of the differentiation process.

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Solution

i. For $y = x^2 e^{-x}$:

1. First Differentiation: We apply the product rule, which states that if $y = u(x)v(x)$, then: $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$

   Here, $u(x) = x^2$ and $v(x) = e^{-x}$.
   Differentiating each part:

   • $u'(x) = 2x$
   • $v'(x) = -e^{-x}$ (because the derivative of $e^{-x}$ is $-e^{-x}$)

   Using the product rule:

$$y_1 = e^{-x} (2x) + x^2 (-e^{-x}) = e^{-x} (2x - x^2)$$

2. Second Differentiation: Now, we differentiate $y_1 = e^{-x} (2x - x^2)$ with respect to $x$.

   Again, applying the product rule:

   • First term: $e^{-x} (2x)$ gives $e^{-x} (2 - 2x)$
   • Second term: $x^2 (-e^{-x})$ gives $-e^{-x} (2x - x^2)$

   Combining these results:

$$y_2 = e^{-x} (2 - 2x) + (2x - x^2)(-e^{-x})$$

Simplifying:

$$y_2 = e^{-x} (x^2 - 4x + 2)$$

Final Answer for part i: $y_2 = e^{-x} (x^2 - 4x + 2)$

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ii. For $y = \ln \left( \frac{2x+3}{3x+2} \right)$:

1. First Differentiation: We use the logarithmic property that $\ln(a/b) = \ln(a) - \ln(b)$, so:

$$y = \ln(2x+3) - \ln(3x+2)$$

Now, differentiating each term:

• Derivative of $\ln(2x+3)$ is $\frac{2}{2x+3}$
• Derivative of $\ln(3x+2)$ is $\frac{3}{3x+2}$

So, the first derivative $y_1$ is:

$$y_1 = \frac{2}{2x+3} - \frac{3}{3x+2}$$

2. Second Differentiation: Now, we differentiate $y_1$ using the quotient rule: $\frac{d}{dx} \left( \frac{a}{b} \right) = \frac{b \cdot a' - a \cdot b'}{b^2}$

   Applying this to both terms:

   • For the first term, we apply the quotient rule:

$$\frac{d}{dx}\left(\frac{2}{2x+3}\right) = \frac{-(2)(2)}{(2x+3)^2} = \frac{-4}{(2x+3)^2}$$

• For the second term, applying the quotient rule again:

$$\frac{d}{dx}\left(\frac{3}{3x+2}\right) = \frac{-(3)(3)}{(3x+2)^2} = \frac{-9}{(3x+2)^2}$$

Therefore, the second derivative $y_2$ is:

$$y_2 = \frac{-4}{(2x+3)^2} + \frac{9}{(3x+2)^2}$$

To simplify, we combine the fractions:

$$y_2 = \frac{-4(3x+2)^2 + 9(2x+3)^2}{(2x+3)^2 (3x+2)^2}$$

Expanding both squares:

$$= \frac{-4(9x^2 + 12x + 4) + 9(4x^2 + 12x + 9)}{(2x+3)^2 (3x+2)^2}$$

After expanding:

$$y_2 = \frac{60x + 65}{(2x+3)^2 (3x+2)^2}$$

Final Answer for part ii: $y_2 = \frac{60x + 65}{(2x+3)^2 (3x+2)^2}$

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Key Formulas or Methods Used

• Product Rule:
  $\frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)$

• Quotient Rule:
  $\frac{d}{dx}\left(\frac{a(x)}{b(x)}\right) = \frac{b(x)a'(x) - a(x)b'(x)}{b(x)^2}$

• Logarithmic Differentiation: $\frac{d}{dx} \ln(u(x)) = \frac{u'(x)}{u(x)}$

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Summary of Steps

1. For part i:

   • Differentiate $y = x^2 e^{-x}$ using the product rule.
   • Differentiate again to find $y_2$.

2. For part ii:

   • Break down $y = \ln \left( \frac{2x+3}{3x+2} \right)$ using logarithmic properties.
   • Differentiate the two logarithmic terms.
   • Apply the quotient rule for the second derivative.
   • Simplify the final expression for $y_2$.