Question

y=12F+Je^(t)

where F and J are given constants.
What is the derivative?

Answer 1

The derivative of a constant =0
hence.
\[y=12F + Je^f \]
yields
\[y'=12F\frac{ d }{ dt } + Je^t \frac{ d }{ dt } = 0+e^t\]

Answer 2

sorry first e^f should be e^t

Also I used chain rule onto the e^t

\[e^t \frac{ d }{ dt } = e^t * t'\]

Answer 3

it isnt working :(

Answer 4

@satellite73

Answer 5

Kainashi is on the right track, but not quite there!

Start with the function:

\[y = 12F+Je^t\]

Now take the derivative of both sides:

\[\frac{dy}{dt} = \frac{d}{dt} ( 12 F +Je^t)\]

The derivative has two important properties that we can use now:

For any two functions, the derivative of the sum is the sum of the derivatives:
\[\frac{d}{dt}( f(t)+g(t)) = \frac{d}{dt}( f(t)) + \frac{d}{dt}( g(t))\]

And the other trick we have is that for any constant C we can pull it outside the derivative, and this is really just a consequence of the product rule since the derivative of a constant by itself is zero!
\[\frac{d}{dt} ( C f(t)) = C \frac{d}{dt}(f(t))\]

So try to use these tricks to solve your problem, give it your best shot.

Answer 6

the constant isnt C though its F and J

Answer 7

Yeah so now you can use these general rules to solve your problem by replacing C and f(t) with your functions.

Answer 8

FJ(12+e^t)?

Answer 9

I got it J(e^t)

Answer 10

Yeah that's the right answer

Answer 11

But what you just wrote right before that is not right, since

\[FJ(12+e^t) \ne F12 + Je^t\]

Answer 12

yeah I know