nth derivative of 10^-x

In general,  (d/dx) (a^x)  = (ln a)* a^x * (the derivative of the exponent on "a").....so......

(d/dx) 10^-x =  ln(10)* 10^-x * (-1)  = -(ln10)10^-x     .....and.....

(d/dx) [- ln(10)* 10^-x ] =  -(ln10) * (ln 10)* 10^-x (-1)  =  (ln10)^2* 10^-x

So.......the "nth" derivative would be

(-1)^n * (ln 10)^n * 10^-x

$$\small{\text{$\mathbf{n^{th}}$ derivative of $\mathbf{10^{-x}}$}}$$

$$\text{ \small{Formula} $
\boxed{
\begin{array}{lcl}
y &=& e^{k\cdot x} \\
y' &=& k \cdot e^{k\cdot x} \qquad y'' = k^2 \cdot e^{k\cdot x} \qquad y''' = k^3 \cdot e^{k\cdot x} \qquad \cdots \\
y^{(n)} & = & k^n \cdot e^{k\cdot x}
\end{array}
}
$} \\\\\\
\small{\text{$
\begin{array}{lcl}
y &=& 10^{-x} \qquad \Rightarrow \qquad
y = e^{-x\cdot \ln{(10)} } = e^{ \overbrace{\left[
-\ln{10} \right] }^{=\big{k}} \cdot x }\\
\end{array}
$}} \\\\
\small{\text{$
\underline{ y^{n} = k^n \cdot e^{k\cdot x} }
\qquad \underline{ k = -\ln{(10)} }
$}} \\\\
\small{\text{$
\begin{array}{lcl}
y^{(n)} &=& [-\ln{(10)} ]^n \cdot
\underbrace{ e^{ [-\ln{(10)} ]\cdot x} }_{=\big{10^{-x}}}\\\\
\mathbf{y^{(n)}} & \mathbf{=} & \mathbf{[-\ln{(10)} ]^n \cdot 10^{-x} }\\\\
\end{array}
$}} \\\\$$