मराठी

√ 2 X 2 + 1

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प्रश्न

 (x2 + 1) (x − 5)

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उत्तर

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

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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 1.13 | पृष्ठ २५

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