Derivative of $1-5^{-x}$

Question

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 $$y' =\frac{x}5 $$

Answer 1

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

$$ f'(x) = - e^{-x \ln {5}} \cdot \left( -x \ln{5} \right)' = -e^{-x \ln {5}} (-\ln{5}) = e^{-x \ln{5}} \ln{5} = 5^{-x} \ln {5}.$$

Answer 2

$$1-5^{-x}$$

$$=(1)'-(5^{-x})'$$

Now use the chain rule:

$\color{gray}{\frac{d}{dx}(5^{-x})=\frac{d5^{\varphi}}{d\varphi}\frac{d\varphi}{dx},\text{where }\varphi=-x\text{ and } \frac{d}{d\varphi}(5^{\varphi})=5^{\varphi}\ln(5)}$

$$=0-5^{-x}\ln(5)(-x)'$$

$$=\boxed{\color{red}{5^{-x}\ln(5)}}$$

Answer 3

Recall that

1. The derivative of a constant is 0, and,
2. 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}

Answer 4

you should use formula $\frac{da^x}{dx}=a^x\ln |a|$ then

$\frac{d}{dx}(1-5^{-x})=\frac{d}{dx}(1)-\frac{d}{dx}(5^{-x})=0-5^{-x}\ln 5(-1)=5^{-x}\ln 5$

Answer 5

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 $$ f'(x) = \ln(1/5)\left(\frac{1}{5}\right)^x = -\ln(5)5^{-x}. $$

Answer 6

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 $1 - y = 5^{-x}$.