+0  
 
0
275
2
avatar

nth derivative of 10^-x

Guest Aug 16, 2015

Best Answer 

 #1
avatar+78744 
+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

 

 

 

CPhill  Aug 16, 2015
Sort: 

2+0 Answers

 #1
avatar+78744 
+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+18715 
+5

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

 

 

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

 

heureka  Aug 17, 2015

5 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details