हिंदी

X + 1 X + 2

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प्रश्न

\[\frac{x + 1}{x + 2}\]

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उत्तर

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

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अध्याय 30: Derivatives - Exercise 30.2 [पृष्ठ २५]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 30 Derivatives
Exercise 30.2 | Q 1.06 | पृष्ठ २५

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