Finding the force of mortality at (x+t) from the force of mortality at x

Question

Let $\mu_x$ be the force of mortality and $S_X(x)$ the survival function

I have the proof that $\mu_x=\frac{-d(ln(S_X(x))}{dx}$

but then it is deduced that $\mu_{x+t}=\frac{-d(ln(S_X(x+t))}{dt}$

Could you please explain me how $dx$ became $dt$?

Thank you

Answer 1

I got the answer in the comment section .

$\mu_x(x)$ is the force of mortality at age x but we want to find $\mu_{x+t}$ in order to find ${t}P_{x}$ = Pr(X>x+t/X>x)$ where X is a r.v representing the age at death.

$\mu_x=\frac{-d(ln(S_X(x))}{dx}$

<=>

$\mu_{x+t}=\frac{-d(ln(S_X(x+t))}{d(x+t)} =\mu_{x+t}=\frac{-d(ln(S_X(x+t))}{d(x)+d(t)} $ but since we are holding the age x and and want to see only the variation of $\mu$ when $t$ varies we put dx=0 and we get the answer.

Very simple and basic but didn't get it without the comments!

Thank you