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प्रश्न
Mark the correct alternative in of the following:
If\[y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + . . .\]then \[\frac{dy}{dx} =\]
विकल्प
y + 1
y − 1
y
y2
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उत्तर
\[y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + . . .\]
Differentiating both sides with respect to x, we get \[\frac{dy}{dx} = \frac{d}{dx}\left( 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + . . . \right)\]
\[ = \frac{d}{dx}\left( 1 \right) + \frac{d}{dx}\left( \frac{x}{1!} \right) + \frac{d}{dx}\left( \frac{x^2}{2!} \right) + \frac{d}{dx}\left( \frac{x^3}{3!} \right) + \frac{d}{dx}\left( \frac{x^4}{4!} \right) + . . . \]
\[ = \frac{d}{dx}\left( 1 \right) + \frac{1}{1!}\frac{d}{dx}\left( x \right) + \frac{1}{2!}\frac{d}{dx}\left( x^2 \right) + \frac{1}{3!}\frac{d}{dx}\left( x^3 \right) + \frac{1}{4!}\frac{d}{dx}\left( x^4 \right) + . . . \]
\[ = 0 + \frac{1}{1!} \times 1 + \frac{1}{2!} \times 2x + \frac{1}{3!} \times 3 x^2 + \frac{1}{4!} \times 4 x^3 + . . . \left( y = x^n \Rightarrow \frac{dy}{dx} = n x^{n - 1} \right)\]
\[= 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + . . . \left[ \frac{n}{n!} = \frac{1}{\left( n - 1 \right)!} \right]\]
\[ = y\]
\[\therefore \frac{dy}{dx} = y\]
Hence, the correct answer is option (c).
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