Question

Mark the correct alternative in of the following:

If\[y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + . . .\]then \[\frac{dy}{dx} =\] 

 

Options

•  y + 1          

• y − 1          

• y   

•  y²

Answer 1

\[y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + . . .\] 

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).