What is the derivative of $f(x)=-e^x$?

Question

I have this function:

$y= (15 - e^{3t})$

I am pretty sure the derivative of that function, using Chain Rule, is

$y'= -e^{3t}*3$

But why is it negative again? Maybe I'm just having a brainfart... but why is it negative? The only rule I know is that the derivative of $y = e^x$ is $e^x$. So given that rule, what is the derivative of $-e^x$?

I remember my limit rules a bit like...

$\lim_{x \to 2} -2x$

$= -1 \lim_{x \to 2} 2x$

$= -4$

Does that concept apply here?

Answer 1

In general, if the derivative of $f(x)$ is $f'(x)$, then the derivative of $-f(x)$ is $-f'(x)$; also, the derivative of $C-f(x)$, for any constant $C$, is $-f'(x)$.

So if we let $C = 15$, and $f(x) = e^{3x}$, then $f'(x) = 3e^{3x}$, and

$$ \frac{d}{dx} 15-e^{3x} = -3e^{3x} $$

Answer 2

We have:

$y = (15 - e^{3t})$

$$y' = \underbrace{0}_{\frac d{dt}(15)} - (\underbrace{3}_{\frac d{dt}(3t)}\cdot e^{3t})$$

You see, $3$ in $3t$ is just a constant. When, via the chain rule, we take $$\frac{d}{dt}\left(15- e^{3t}\right) = \frac d{dt} (15) -\frac d{dt}(e^{3t}) =0 - \left(e^{3t} \cdot \frac d{dt} (3t)\right) = - \left( e^{3t}\cdot 3\right) = -3\left( e^{3t}\right)$$

In either case, either $15- e^{et}$ or $15-e^{3t}$, or $15-e^{1231791\, t}$, or $15 - e^{kt}$ where $k$ is any real constant, we get that the derivative is equal to $-k\cdot e^{kt}$, in each case, by the chain rule.

---

$−e^{kt}$ is to be treated just as we would treat $−(e^{kt}).$ It is not to be taken to mean $(−e)^{kt}$. That's a major difference.

In general, when we have $e^{f(x)}$, the derivative is $e^{f(x)}\cdot f'(x)$

Answer 3

Is your function $(15-e^{3t})$? Because you said you have a function $(15-e^{et})$ but you have derivative $-e^{3t}$. If your fcn is $(15-e^{3t})$, Its derivative is $-e^{3t}$ *3. Dont forget that $e$ is negative so you get a negative answer

Answer 4

The derivative of 15 is 0 and you are left with finding the derivative of -e^3t, imagine the equation like this (-1)*(e^3t) and now simply factor out the (-1) constant, apply derivative and factor the constant back in.