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

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

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