Question

find dy/dx pf the following y=1+x/1!+x²/2!+x³/3!+...​

Answer 1

Answer : y

Step-by-step explanation:

$y=1+x +\frac{x^{2} }{2!}+ \frac{x^{3} }{3!}+ \frac{x^{4} }{4!} +\frac{x^{5} }{5!} ...\\\frac{dy}{dx} =0+1+\frac{2x} {2!} + \frac{3x }{3!}+ \frac{4x }{4!} +\frac{5x}{5!} ...\\\frac{dy}{dx}=1+\frac{2x} {2*1!} + \frac{3x }{3*2!}+ \frac{4x }{4*3!} +\frac{5x}{5*4!} ...\\\frac{dy}{dx}=1+x +\frac{x^{2} }{2!}+ \frac{x^{3} }{3!}+ \frac{x^{4} }{4!} +\frac{x^{5} }{5!} ...$

Hence ,

$\frac{dy}{dx}=y$