How come this derivative is true?

Question

I am taking the derivative of forward priced security.

$$ \text{Forward Price} = S\mathrm{e}^{r(T-t)} $$ Thus, $$ \frac{d}{dS}\text{Forward Price} = \mathrm{e}^{r(T-t)} $$ but I don't get the derivative with respect to $t$ implies $\frac{d}{dt} \text{Forward Price} = -r\mathrm{e}^{r(T-t)}$

How come we have a negative $r$ in front now?

Answer 1

I'll write this out with full details. You are trying to take the derivative with respect to $t$ of a function of the form $ce^{f(t)}$ where $c$ is a constant. In this case, $c=S$ and $f(t)$ is $r(T-t)$. $\dfrac{d}{dt} ce^{f(t)} = c \dfrac{d}{dt} e^{f(t)} = c f'(t) e^{f(t)}$. In this case $f(t)=r(T-t)$, which can be written as $rT-rt$ so $f'(t)=-r$.

Answer 2

We have price $P =Se^{r(T-t)}$. Let the term $r(T-t) =\alpha $ for now. We know that the derivative of $e^x $ with respect to $x $ is also $e^x $. But here $\alpha $ is a function of $t $, so we need to additionally apply the chain rule. Hence, $$\frac {dP}{dt} =\frac {d}{dt}(Se^{r(T-t)}) =Se^{r(T-t)} \times \frac {d}{dt}(r(T-t)) = Se^{r(T-t)}\cdot (-r) $$ Hence, we have got the derivative as required. Hope it helps.