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nth derivative of 10^-x

 Aug 16, 2015

Best Answer 

 #1
avatar+94545 
+10

In general,  (d/dx) (ax)  = (ln a)* ax * (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

 

 

 

 Aug 16, 2015
 #1
avatar+94545 
+10
Best Answer

In general,  (d/dx) (ax)  = (ln a)* ax * (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

 

 

 

CPhill Aug 16, 2015
 #2
avatar+20831 
+5

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

 

 

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

 

 Aug 17, 2015

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