Derivative of $1-5^{-x}$ What is the derivative of $y=1-5^{-x}$.
Any help is greatly appreciated!
I have tried using logs, but I don't think it is correct;
$$y=1-5^{-x}$$
$$\ln(y)=\ln(1) +x\ln(5)$$
$$y =x\ln(5)$$
and hence
$$y' =\frac{x}5 $$
 A: You can solve the problem using the idea of logs but note that the equality you wrote is wrong. You have
$$ f(x) = 1 - 5^{-x} = 1 - e^{\ln(5^{-x})} = 1 - e^{-x \ln{5}} $$
and so using the chain rule you have
$$ f'(x) = - e^{-x \ln {5}} \cdot \left( -x \ln{5} \right)' = -e^{-x \ln {5}} (-\ln{5}) = e^{-x \ln{5}} \ln{5} = 5^{-x} \ln {5}.$$
A: 
$$1-5^{-x}$$

$$=(1)'-(5^{-x})'$$
Now use the chain rule:
$\color{gray}{\frac{d}{dx}(5^{-x})=\frac{d5^{\varphi}}{d\varphi}\frac{d\varphi}{dx},\text{where   }\varphi=-x\text{  and  } \frac{d}{d\varphi}(5^{\varphi})=5^{\varphi}\ln(5)}$
$$=0-5^{-x}\ln(5)(-x)'$$
$$=\boxed{\color{red}{5^{-x}\ln(5)}}$$
A: Recall that 


*

*The derivative of a constant is 0, and, 

*The derivative of a sum is the sum of the derivative, then


\begin{align}
f'(x) &= (1)' - (5^{-x})' \\
&= -(e^{-x\ln(5)})' \\
&= -(e^{-x\ln(5)})(-\ln(5)) \\
&= (e^{-x\ln(5)})(\ln(5)) \\
&= (5^{-x})(\ln(5))
\end{align}
A: you should use formula $\frac{da^x}{dx}=a^x\ln |a|$ then
$\frac{d}{dx}(1-5^{-x})=\frac{d}{dx}(1)-\frac{d}{dx}(5^{-x})=0-5^{-x}\ln 5(-1)=5^{-x}\ln 5$
A: You need to use the rule that if 
$$
f(x) = a^x\quad\quad (a>0)
$$
then
$$
f'(x) = \ln(a)a^x.
$$
Now you have the function $5^{-x}$, but this is just
$$
5^{-x} = \left(\frac{1}{5}\right)^x.
$$
So if $f(x) = 5^{-x}$, then
$$
f'(x) = \ln(1/5)\left(\frac{1}{5}\right)^x = -\ln(5)5^{-x}.
$$
A: If the rule for differentiating $a^x$ confuses you, you can rearrange for $x$ instead:
$$1 - y = 5^{-x}$$
$$-\frac{1}{\ln 5} \ln(1 - y) = x$$
and now using implicit differentiation:
$$-\frac{1}{\ln 5} \frac{-1}{1 - y} \frac{dy}{dx} = 1$$
$$\frac{dy}{dx} = (\ln 5) (1-y)$$
$$=(\ln 5) \ 5^{-x}$$
as we earlier said that $1 - y = 5^{-x}$.
