First slide

Derivatives

Question

$If y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots . + \frac{x^{n}}{n!} then \frac{d y}{d x} - y + \frac{x^{n}}{n!} =$

Moderate

A

1
B

0
C

-1
D

None

Solution

$\begin{array}{l} \frac{d y}{d x} = 0 + \frac{1}{1!} + \frac{1}{2!} 2 x + \frac{1}{3!} 3 x^{2} + \dots . + \frac{1}{n!} (n x^{n - 1}) \\ = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots \cdot + \frac{x^{n - 1}}{(n - 1)!} \\ = \{1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots . + \frac{x^{n - 1}}{(n - 1)!} + \frac{x^{n}}{n!}\} - \frac{x^{n}}{n!} \\ \frac{d y}{d x} = y - \frac{x^{n}}{n!} \\ \frac{d y}{d x} - y + \frac{x^{n}}{n!} = 0 \end{array}$

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