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प्रश्न
Differentiate the following functions from first principles sin−1 (2x + 3) ?
उत्तर
\[\text{Let} f\left( x \right) = \sin^{- 1} \left( 2x + 3 \right)\]
\[ \Rightarrow f\left( x + h \right) = \sin^{- 1} \left( 2\left( x + h \right) + 3 \right)\]
\[ \Rightarrow f\left( x + h \right) = \sin^{- 1} \left( 2x + 2h + 3 \right)\]
\[ \therefore \frac{d}{dx}\left\{ f\left( x \right) \right\} = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \left( 2x + 2h + 3 \right) - \sin^{- 1} \left( 2x + 3 \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \left[ \left( 2x + 2h + 3 \right)\sqrt{1 - \left( 2x + 3 \right)^2} - \left( 2x + 3 \right)\sqrt{1 - \left( 2x + 2h + 3 \right)^2} \right]}{h} \left[ \because \sin^{- 1} x - \sin^{- 1} y = \sin^{- 1} \left[ x\sqrt{1 - y^2} - y\sqrt{1 - x^2} \right] \right]\]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} z}{z} \times \frac{z}{h}\]
\[\text{where, } z = \left( 2x + 2h + 3 \right)\sqrt{1 - \left( 2x + 3 \right)^2} - \left( 2x + 3 \right)\sqrt{1 - \left( 2x + 2h + 3 \right)^2} \text{ and }\lim_{h \to 0} \frac{\sin^{- 1} h}{h} = 1\]
\[ = \lim_{h \to 0} \frac{z}{h}\]
\[ = \lim_{h \to 0} \frac{\left( 2x + 2h + 3 \right)\sqrt{1 - \left( 2x + 3 \right)^2} - \left( 2x + 3 \right)\sqrt{1 - \left( 2x + 2h + 3 \right)^2}}{h}\]
\[ = \lim_{h \to 0} \frac{\left( 2x + 2h + 3 \right)^2 \left\{ 1 - \left( 2x + 3 \right)^2 \right\} - \left( 2x + 3 \right)^2 \left\{ 1 - \left( 2x + 2h + 3 \right)^2 \right\}}{h\left\{ \left( 2x + 2h + 3 \right)\sqrt{1 - \left( 2x + 3 \right)^2} + \left( 2x + 3 \right)\sqrt{1 - \left( 2x + 2h + 3 \right)^2} \right\}} \] ...........[Rationalizing numerator]
\[ = \lim_{h \to 0} \frac{\left[ \left( 2x + 3 \right)^2 + 4 h^2 + 4h\left( 2x + 3 \right) \right]\left\{ 1 - \left( 2x + 3 \right)^2 \right\} - \left( 2x + 3 \right)^2 \left[ 1 - \left( 2x + 3 \right)^2 - 4 h^2 - 4h\left( 2x + 3 \right) \right]}{h\left\{ \left( 2x + 2h + 3 \right)\sqrt{1 - \left( 2x + 3 \right)^2} + \left( 2x + 3 \right)\sqrt{1 - \left( 2x + 2h + 3 \right)^2} \right\}}\]
\[ = \lim_{h \to 0} \frac{\left[ \left( 2x + 3 \right)^2 + 4 h^2 + 4h\left( 2x + 3 \right) - \left( 2x + 3 \right)^4 - 4 h^2 \left( 2x + 3 \right)^2 - 4h \left( 2x + 3 \right)^3 - \left( 2x + 3 \right)^2 + \left( 2x + 3 \right)^4 + 4 h^2 \left( 2x + 3 \right)^2 + 4h \left( 2x + 3 \right)^3 \right]}{h\left\{ \left( 2x + 2h + 3 \right)\sqrt{1 - \left( 2x + 3 \right)^2} + \left( 2x + 3 \right)\sqrt{1 - \left( 2x + 2h + 3 \right)^2} \right\}}\]
\[ = \lim_{h \to 0} \frac{4h\left[ h + \left( 2x + 3 \right) \right]}{h\left\{ \left( 2x + 2h + 3 \right)\sqrt{1 - \left( 2x + 3 \right)^2} + \left( 2x + 3 \right)\sqrt{1 - \left( 2x + 2h + 3 \right)^2} \right\}}\]
\[ = \frac{4\left( 2x + 3 \right)}{\left( 2x + 3 \right)\sqrt{1 - \left( 2x + 3 \right)^2} + \left( 2x + 3 \right)\sqrt{1 - \left( 2x + 3 \right)^2}}\]
\[ = \frac{4\left( 2x + 3 \right)}{2\left( 2x + 3 \right)\sqrt{1 - \left( 2x + 3 \right)^2}}\]
\[ = \frac{2}{\sqrt{1 - \left( 2x + 3 \right)^2}}\]
\[ \therefore \frac{d}{dx}\left\{ \sin^{- 1} \left( 2x + 3 \right) \right\} = \frac{2}{\sqrt{1 - \left( 2x + 3 \right)^2}}\]
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