Why is the derivative multiplied by $-1$?

Question

I have the step by step answers, but I can't discover where the "$-1$" comes from. A constant $= 0$, so it can't be from the inside function.

$$\begin{align*}P'(t) &= \frac{d}{dt}\left[20\left(1+e^{10-t}\right)^{-1}\right] \\ &= -20\left(1+e^{10-t}\right)^{-2}\cdot \frac{d}{dt}\left[1+e^{10-t}\right] \\ &= -20\left(1+e^{10-t}\right)^{-2}\cdot \color{red}{\boxed{\color{black}{e^{10-t}\cdot (-1)}}} \\ &= 20e^{10-t}\left(1+e^{10-t}\right)^{-2}\end{align*} \\$$

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Thank you so much for the quick responses.

Answer 1

Because $e^{10-t}$ has also to be differentiated using the chain rule:

• $\frac d{dt}(e^{10-t}) = e^{10-t}\cdot \frac d{dt}(10-t) = e^{10-t}\cdot (-1)$

Answer 2

The derivative is multiplied by $-1$ because that is the derivative of the exponent, $10-t$.

Answer 3

In general, the derivative of an exponential function is given by

$$\frac{d}{dt} e^{f(t)} = e^{f(t)} \cdot \frac{d}{dt}\left(f(t))\right).$$

Here, we can identify $f(t) = 10 - t$, so we obtain

$$\frac{d}{dt} e^{10 - t} = e^{10 - t} \cdot \frac{d}{dt} (10 - t) = -e^{10 - t}. $$

Answer 4

$\frac{d}{dt} e^{10-t}=\frac{d}{dt}(10-t)\times e^{10-t}=(-1)\times e^{10-t}$