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

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

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

2 years ago

What does h(40)=1820 mean it terms of problem?!!! HELPP

mylen [45]

Answer:

D is the correct option.

Step-by-step explanation:

In the given table, the first column is for "hours of training" and the second column is for "Monthly pay".

Now, we have to state the meaning of the term h(40)=1820.

In coordinate system we can write this as (40,1820).

Since, the first column represents "hours of training" and second column is for "Monthly pay". Hence, we can say that for 40 hours of training the monthly payment is $1820.

In other words, if the training attended by the worker is 40 hours, h/she would pay $1820.

Therefore, D is the correct option.

5 0

3 years ago

6 - 2 to the third power + (-9 + 5) · 2.

Alexxx [7]

Answer:

Step-by-step explanation:

6- 2 = 4 / 4*4*4 = 32

-9+5=4/4*2=8

32 + 8 = 40

Could really use brainliest if thif helps