Question

If $\rm y=1+\frac{x}{1!}+\frac{x^{2}}{2!} +\frac{x^{3}}{3!}+...$ find $\rm \frac{dy}{dx}$.

Answer 1

it is given that, y = 1 + x/1! + x²/2! + x³/3! + .....

we have to find dy/dx

we know, 1! = 1,

2! = 2 × 1 = 2

3! = 3 × 2 × 1 = 6,

4! = 4 × 3 × 2 × 1 = 24

5! = 5 × 4 × 3 × 2 × 1 = 120

..... ...... ....

now, y = 1 + x + x²/2 + x³/6 + x⁴/24 + ...

.....

differentiating both sides,

dy/dx = d(1)/dx + d(x)/dx + 1/2 × d(x²)/dx + 1/6 × d(x³)/dx + 1/24 × d(x⁴)/dx + ....

= 0 + 1 + 1/2 × 2x + 1/6 × 3x² + 1/24 × 4x³ + .....

= 1 + x + x²/2 + x³/6 + .......

now, dy/dx = 1 + x + x²/2 + x³/6 + .......

= 1 + x/1! + x²/2! + x³/3! + ......

= y

hence, dy/dx = y

[note : according to binomial expansion, e^x = 1 + x/1! + x²/2! + x³/3! + ....

so, y = e^x and we know, differentiation of e^x = e^x

that's why dy/dx = y ]

Answer 2

Answer:

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

Step-by-step explanation:

In this question

We have been given that, $y\ =\ 1\ +\ \frac{x}{1!}\ +\ \frac{x^{2}}{2!}\ +\ \frac{x^{3}}{3!}\ +\ .....$

We need to find $\frac{dy}{dx}$

We know that 1! = 1,

2! = 2 × 1 = 2

3! = 3 × 2 × 1 = 6,

4! = 4 × 3 × 2 × 1 = 24,

5! = 5 × 4 × 3 × 2 × 1 = 120,

......

......

Now, $y\ =\ 1\ +\ x\ +\ \frac{x^{2}}{2}\ +\ \frac{x^{3}}{6}\ +\ \frac{x^{4}}{24}\ +...$

On differentiating both the sides we get,

$\frac{dy}{dx}\ =\ \frac{d(1)}{dx}\ +\ \frac{d(x)}{dx}\ +\ \frac{1}{2}\times \frac{d(x^{2})}{dx}\ +\ \frac{1}{6}\times \frac{d(x^{3})}{dx}\ +\ \frac{1}{24}\times \frac{d(x^{4})}{dx} + ....$

= $0\ +\ 1\ +\ \frac{1}{2}\times 2x + \frac{1}{6}\times 3x^{2} + \frac{1}{24}\times 4x^{3} + .....$

= $1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + .......$

Now, it is given that $\frac{dy}{dx}\ =\ 1\ + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + .......$

= $1 + \frac{x}{1!} + \frac{x^{2}}{2!}\ + \frac{x^{3}}{3!} + ......$

= y

Hence, $\mathbf{\frac{dy}{dx} = y}$