We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
723
2
avatar

nth derivative of 10^-x

 Aug 16, 2015

Best Answer 

 #1
avatar+100529 
+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+100529 
+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+22182 
+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

26 Online Users

avatar
avatar