Question

Differentiate the following function:
y=7K+Me^t,

where K and M are given constants.
Enter the derivative here:

Answer 1

@greatlife44

Answer 2

hi again :D

Answer 3

hi :)

Answer 4

\[y = 7K + Me^{t}\]

so remember, the derivative of a constant is just going to be zero, because think about it there isn't any variable so the slope of a constant is going to be zero

\[\frac{ d }{ dx }c = 0 \]

Answer 5

now let's go on to this.

the derivative of e^{t} is going to just be e^{t}

\[\frac{ d }{ dx } e^{t} = e^{t}\]

now let's apply these rules to our question.

\[\frac{ dy }{ dt }7K + \frac{ dy }{ dt } Me^{t} = 0 + Me^{t}\]

\[\frac{ dy }{ dt } = Me^{t}\]

Answer 6

but why wouldn't M be zero as well

Answer 7

so when we say

\[\frac{ dy }{ dt }\] this means we are taking the derivative of y with respect to t

Answer 8

oh okay

Answer 9

think of it like this

\[\frac{ dy }{ dt } Me^{t} = M \frac{ dy }{ dt }e^{t}\]

remember the derivative of e^{t} is just going to be e^{t}

Answer 10

@Pulsified333 try this

find the derivative of

\[2x \]

Answer 11

2

Answer 12

okay,

Answer 13

so we know that the slope of a line is going to be constant

Answer 14

right

Answer 15

y = 2

what about the derivative of this?

Answer 16

that's not the same as if we took the derivative of say

\[y = 2e^{x}\]

in this case 2 is your constant.

Answer 17

because remember if 2 which is our constant = 0 then the derivative would be zero because remember

\[\frac{ d }{ dx } 2e^{x} = 2e^{x}, 2*e^{x}, so~our~constant~can't~be~ zero. \]

Answer 18

true

Answer 19

think about it @Pulsified333

\[\frac{ d }{ dx } Ae^{x} = Ae^{x}\]

Answer 20

in this case our constant is not zero, because then it would mean 0*e^{x} which would just be zero. I guess there's a better way to explain it but yeah I guess that's the idea.

Answer 21

I get it now thank you :D

Answer 22

great :)