मराठी

X + 2 3 X + 5

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प्रश्न

\[\frac{x + 2}{3x + 5}\]

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उत्तर

\[ \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\frac{x + h + 2}{3\left( x + h \right) + 5} - \frac{x + 2}{3x + 5}}{h}\]
\[ = \lim_{h \to 0} \frac{\frac{x + h + 2}{3x + 3h + 5} - \frac{x + 2}{3x + 5}}{h}\]
\[ = \lim_{h \to 0} \frac{\left( x + h + 2 \right)\left( 3x + 5 \right) - \left( 3x + 3h + 5 \right)\left( x + 2 \right)}{h\left( 3x + 3h + 5 \right)\left( 3x + 5 \right)}\]
\[ = \lim_{h \to 0} \frac{3 x^2 + 3xh + 6x + 5x + 5h + 10 - 3 x^2 - 3xh - 5x - 6x - 6h - 10}{h\left( 3x + 3h + 5 \right)\left( 3x + 5 \right)}\]
\[ = \lim_{h \to 0} \frac{- h}{h\left( 3x + 3h + 5 \right)\left( 3x + 5 \right)}\]
\[ = \lim_{h \to 0} \frac{- 1}{\left( 3x + 3h + 5 \right)\left( 3x + 5 \right)}\]
\[ = \frac{- 1}{\left( 3x + 5 \right)^2}\]

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पाठ 30: Derivatives - Exercise 30.2 [पृष्ठ २५]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 30 Derivatives
Exercise 30.2 | Q 1.07 | पृष्ठ २५

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