What do I have to do different when taking derivative of $2^{-x}$ opposed to $2^x$ So for $2^x$ I know the derivative would be
$$2^x \ln(2)$$
What would be the thing or step I'd do different for $2^{-x}$ ? 
I've trying to take the derivative of  
$$40\over1+2^{-t}$$ 
& I keep getting the wrong asnwer, I think it's because I'm not sure about how to handle the $2^{-t}$
 A: The way of doing this I am familiar with is:
$2^{-x}= e^{-xln2}$ , and then you find $\frac {d}{dx} e^{-xln2}$ using the chain rule,
so that $\frac{d}{dt}e^{-tln2}=e^{tln2} \frac{d}{dt}(-tln2)$.
Can you take it from there?
EDIT: The chain rule together with the quotient rule applied here comes down to:
$\frac {d}{dt} \frac {40}{1+2^{-t}}= \frac {0-40(\frac{d}{dt}(1+ 2^{-t})}{(1+2^{-t})^2}$ , where we use $\frac {d}{dt}(40)=0$.
A: I think you might be confused about the derivative of $2^x$ - or you've phrased things weirdly. In any case, the derivative of $2^x=\ln(2) 2^x $. You can find the derivative of $2^{-x}$ in two ways; either you can apply the chain rule to get $-1\cdot\ln(2) 2^{-x}$, where the $-1$ term comes from $-x$. You can also change $2^{-x}=\left(\frac{1}2\right)^x$ and get the derivative as $\ln\left(\frac{1}2\right)\left(\frac{1}2\right)^x$, which is equivalent.
A: If you have:
$$2^{f(x)}$$ you can use the chain rule to find that:
$$\frac{\text{d}}{\text{d}x}2^{f(x)}=f'(x)\times {2^{f(x)}\ln(2)}$$
as others have done you could show this by:
$$\frac{\text{d}}{\text{d}x}2^{f(x)}=\frac{\text{d}}{\text{d}x}e^{\ln(2)\times f(x)}= f'(x)\times\ln(2)\times e^{\ln(2)\times f(x)}= f'(x)\times\ln(2)\times2^{f(x)}$$
then if you let $f(x)=-x$ then you'd have $f'(x)=-1$ so:
$$\frac{\text{d}}{\text{d}x}2^{-x}=-1\times2^{-x}\ln(2)=-2^{-x}\ln(2)$$

thus you just need to apply the quotient rule:
$$\frac{\text{d}}{\text{d}x}\frac{40}{1+2^{-t}}=\frac{0\times(1+2^{-t})-(-2^{-t}\ln(2)\times40)}{(1+2^{-t})^2}=\frac{-(-2^{-t}\ln(2)\times40)}{(1+2^{-t})^2}=\frac{2^{-t}\ln(2)\times40}{(1+2^{-t})^2}$$
for your books solution:
$$\frac{2^{-t}\ln(2)\times40}{(1+2^{-t})^2}=\frac{2^{-t}\ln(2)\times2^2\times 10}{(1+2^{-t})^2}=\frac{2^{-t}\ln(2)\times2^3\times 5}{(1+2^{-t})^2}=\frac{5\times2^{3-t}\ln(2)}{(1+2^{-t})^2}$$
which you can get to Wolfram Alpha's answer by:
$$\frac{5\times2^{3-t}\ln(2)}{(1+2^{-t})^2}=\frac{5\times2^{3-t}\ln(2)}{1+2\times2^{-t}+2^{-2t}}\times \frac{2^{2t}}{2^{2t}}=\frac{5\times2^{3+t}\ln(2)}{1+2\times2^{t}+2^{2t}}=\frac{5\times2^{3+t}\ln(2)}{(1+2^{t})^2}$$
A: The derivative of $y_1 = 2^x$ is $y_1' = \ln 2 . 2^x$; if $y_2 = 2^{-x}$ then $y_2' = -\ln 2.2^{-x}$. 
Hence the derivative of 
$$g(t) = \frac{40}{1 + 2^{-t}}$$
is
$$g'(t) = \frac{40.\ln 2.2^{-t}}{(1 + 2^{-t})^2}$$
A: First of all, the derivative of $2^x = e^{(\ln 2) x}$ is $(\ln 2) e^{(\ln 2) x} = (\ln 2)2^x$, not $1 / ((\ln 2) 2^x)$ (which admittedly was given in an earlier revision of the question).  This follows from the general formula
$\dfrac{de^{u(x)}}{dx} = e^{u(x)}\dfrac{du(x)}{dx} \tag{1}$
by taking $u(x) = (\ln 2) x$; we also used the identity $a = e^{\ln a}$, $a > 0$ to write $2^x= e^{(\ln 2) x}$.  The derivative of $2^{-x}$ can be computed in essentially the same manner; we simply note that $2^{-1} = e^{\ln 2^{-1}} = e^{-\ln 2}$ and set $u(x) = -(\ln 2)x$ to obtain
$\dfrac{d2^{-x}}{dx} = \dfrac{de^{(-\ln 2)x}}{dx} = (-\ln 2)e^{(-\ln 2)x} = (-\ln 2) 2^{-x}. \tag{2}$
As for
$\dfrac{40}{1 + 2^{-t}} = 40(1 + 2^{-t})^{-1}, \tag{3}$
we simply apply the chain rule (which has already occured in the formula (1)) a couple of times, first to obtain
$\dfrac{d}{dt}(40(1 + 2^{-t})^{-1}) = -40(1 + 2^{-t})^{-2}\dfrac{d2^{-t}}{dt}, \tag{4}$
then once again to $d2^{-t}/dt$, which we basically have already done in (2):
$\dfrac{d}{dt}(40(1 + 2^{-t})^{-1}) = -40(1 + 2^{-t})^{-2}(-\ln 2)2^{-t} = \dfrac{40 (\ln 2)2^{-t}}{(1 + 2^{-t})^2} = \dfrac{40 (\ln 2)}{(2^{2t} + 2^t)^2}.  \tag{5}$
And that's as far as I'll take it.  I can't see the advantage to fiddling around too much with powers of $2$ and so forth, or why Steve Wolfram's engine gives a better formula!
Hope this helps.  Cheers!
and as ever,
Fiat Lux!!!
