Let \( y = \log_e \left( \frac{1 - x^2}{1 + x^2} \right), -1 < x < 1 \). Then at \( x = \frac{1}{2} \), the value of \( 225(y' - y'') \) is equal to
- 732
- 746
- 742
- 736
The Correct Option is D
Approach Solution - 1
To solve the problem given, \( y = \log_e \left( \frac{1 - x^2}{1 + x^2} \right) \), we need to find the value of \( 225(y' - y'') \) at \( x = \frac{1}{2} \).
Let's first compute the first derivative, \( y' \), of the given function.
The function can be rewritten using properties of logarithms:
\(y = \log_e(1-x^2) - \log_e(1+x^2)\)
Differentiating with respect to \(x\), we have:
\(y' = \frac{d}{dx}[\log_e(1-x^2) - \log_e(1+x^2)]\)
Using the derivative of the natural logarithm function, we find:
\(y' = \frac{-2x}{1-x^2} - \frac{2x}{1+x^2}\)
Combining the fractions, we get:
\(y' = -2x \left(\frac{1}{1-x^2} + \frac{1}{1+x^2}\right)\)
Using a common denominator, this becomes:
\(y' = -2x \left(\frac{1+x^2 + 1-x^2}{(1-x^2)(1+x^2)}\right)\)
Simplifying:
\(y' = -2x \left(\frac{2}{1-x^4}\right) = \frac{-4x}{1-x^4}\)
Next, we find the second derivative, \( y'' \).
Taking the derivative of \( y' \):
\(y'' = \frac{d}{dx}\left(\frac{-4x}{1-x^4}\right)\)
Using the quotient rule, where \( u = -4x \) and \( v = 1 - x^4 \), gives:
\(y'' = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\)
Where:
\(\frac{du}{dx} = -4\) and \(\frac{dv}{dx} = -4x^3\)
Substituting these, we have:
\(y'' = \frac{(1-x^4)(-4) - (-4x)(-4x^3)}{(1-x^4)^2}\)
Simplifying further:
\(y'' = \frac{-4(1-x^4) - 16x^4}{(1-x^4)^2}\)
\(= \frac{-4 + 4x^4 - 16x^4}{(1-x^4)^2}\)
\(= \frac{-4 - 12x^4}{(1-x^4)^2}\)
Now, we calculate both \( y' \) and \( y'' \) at \( x = \frac{1}{2} \).
\(y'\left(\frac{1}{2}\right) = \frac{-4(\frac{1}{2})}{1-(\frac{1}{2})^4} = \frac{-2}{1-\frac{1}{16}}\)
\(= \frac{-2}{\frac{15}{16}} = \frac{-32}{15}\)
\(y''\left(\frac{1}{2}\right) = \frac{-4 - 12(\frac{1}{2})^4}{(1-(\frac{1}{2})^4)^2}\)
\(= \frac{-4 - \frac{12}{16}}{\left(\frac{15}{16}\right)^2}\)
\(= \frac{-4 - \frac{3}{4}}{\frac{225}{256}}\)
\(= \frac{-\frac{19}{4}}{\frac{225}{256}} = \frac{-19 \cdot 256}{4 \cdot 225}\)
\(= \frac{-19 \cdot 64}{225}\)
Finally, compute \( 225(y' - y'') \) at \( x = \frac{1}{2} \):
\(225 \left(\frac{-32}{15} - \frac{-19 \cdot 64}{225}\right)\)
Combining and simplifying the terms inside the parenthesis:
\(= 225 \left(\frac{-32 \times 15 - (-19 \times 64)}{225}\right)\)
\(= 225 \left(\frac{-480 + 1216}{225}\right)\)
\(= 225 \left(\frac{736}{225}\right)\)
\(= 736\)
Thus, the value of \( 225(y' - y'') \) is 736.
Approach Solution -2
Given: \(y = \log_e \left( \frac{1 - x^2}{1 + x^2} \right)\)
Step 1. First derivative \( \frac{dy}{dx} \):\(\frac{dy}{dx} = y' = \frac{-4x}{1 - x^4}\)
Step 2. Second derivative \( \frac{d^2y}{dx^2} \):\(y'' = \frac{-4(1 + 3x^4)}{(1 - x^4)^2}\)
Step 3. Calculate \( y' - y'' \) at \( x = \frac{1}{2} \): \(y' - y'' = \frac{-4x}{1 - x^4} + \frac{4(1 + 3x^4)}{(1 - x^4)^2}\)
Step 4. Substitute \( x = \frac{1}{2} \) and simplify to find \( 225(y' - y'') \): After calculations: \(225(y' - y'') = 736\)
The Correct Answer is:736
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