$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

Which expression is equivalent to (xy)z? o (x+y)+z O2z(xy) O x(y+z)

____ [38]

Answer:

the first one

Step-by-step explanation:

7 0

3 years ago

Add: (-3p6 + 5) + 9p6

timama [110]

Answer:

6p^6+ 5

Step-by-step explanation:

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((0 - (3 • (p6))) + 5) + 32p6

Step 2 :

Equation at the end of step 2 :

((0 - 3p6) + 5) + 32p6

Step 3 :

Trying to factor as a Sum of Cubes :

3.1 Factoring: 6p6+5

4 0

3 years ago

15 points please help

podryga [215]

Answer:

The range is y> 9

Step-by-step explanation:

8 0

3 years ago