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If Y = Log E ( X a + B X ) X Then X3 Y2 = (A) (Xy1 − Y)2 (B) (1 + Y)2 - Mathematics

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प्रश्न

If \[y = \log_e \left( \frac{x}{a + bx} \right)^x\] then x3 y2 =

 

विकल्प

  • (xy1 − y)2

  • (1 + y)2

  • \[\left( \frac{y - x y_1}{y_1} \right)^2\]

  • none of these

MCQ

उत्तर

(a) (xy1 − y)2

Here,

\[y = \log_e \left( \frac{x}{a + bx} \right)^x \]

\[ \Rightarrow y = x \log_e \left( \frac{x}{a + bx} \right) \]

\[ \Rightarrow y_1 = \log_e \left( \frac{x}{a + bx} \right) + x \times \frac{a + bx}{x}\left( \frac{1}{a + bx} - \frac{bx}{\left( a + bx \right)^2} \right)\]

\[ \Rightarrow y_1 = \log_e \left( \frac{x}{a + bx} \right) + \left( \frac{a}{a + bx} \right) . . . \left( 1 \right)\]

\[ \Rightarrow y_1 = \frac{y}{x} + \left( \frac{a}{a + bx} \right) \left[ \because y = x \log_e \left( \frac{x}{a + bx} \right) \right]\]

\[ \Rightarrow \frac{x y_1 - y}{x} = \frac{a}{a + bx} . . . \left( 2 \right)\]

\[\text{Differentiating } \left( 1 \right) \text { we get }, \]

\[ y_2 = \frac{a + bx}{x}\left( \frac{a + bx - bx}{\left( a + bx \right)^2} \right) - \frac{ba}{\left( a + bx \right)^2}\]

\[ \Rightarrow y_2 = \frac{a}{x\left( a + bx \right)} - \frac{ba}{\left( a + bx \right)^2}\]

\[ \Rightarrow y_2 = \frac{a\left( a + bx \right) - abx}{x \left( a + bx \right)^2}\]

\[ \Rightarrow y_2 = \frac{a^2}{x \left( a + bx \right)^2}\]

\[ \Rightarrow y_2 = \frac{\left( x y_1 - y \right)^2}{x^3} \left[ \text { Using }
\left( 2 \right) \right]\]

\[ \Rightarrow x^3 y_2 = \left( x y_1 - y \right)^2 \]

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अध्याय 12: Higher Order Derivatives - Exercise 12.3 [पृष्ठ २४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 12 Higher Order Derivatives
Exercise 12.3 | Q 20 | पृष्ठ २४

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