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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 $2^{-t}$

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    $\begingroup$ I can never remember this stuff, so I have to figure it out each time. For the derivative of $2^{-x}$, note that the function is $e^{-(\ln 2) t}$. So by the Chain Rule the derivative is $(-\ln 2)e^{-(\ln 2) t}$ or equivalently $(-\ln 2)2^{-t}$. $\endgroup$ Nov 29, 2014 at 0:32
  • $\begingroup$ @AndréNicolas: Boy, truer words were never written! Cheers! $\endgroup$ Nov 29, 2014 at 2:48

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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 $\frac {d}{dt}(40)=0$.

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  • $\begingroup$ $d\over dx$ $e^{-xln2}$ = $-xln2 * e^{-xln2}$ and because $2^{-x}$ = $e^{-xln2}$ then, I should get $-xln2 * 2^{-x}$ ? $\endgroup$
    – Nolohice
    Nov 29, 2014 at 0:56
  • $\begingroup$ @stunstool: You should get $ln2*2^{-x}$, since $\frac{d}{dx} e^{-xln2}=e^{-xln2}*\frac{d}{dx}(-xln2)=2^{-x}*ln2$. $\endgroup$
    – user192680
    Nov 29, 2014 at 1:16
  • $\begingroup$ when I do $d\over dx$ $(-xln2)$ I get $-ln2$ Which I guess means then I'd end up with $-ln2 * 2^{-x}$ $\endgroup$
    – Nolohice
    Nov 29, 2014 at 1:24
  • $\begingroup$ Yes, sorry, should be -ln2, that is correct, and that is the right answer. $\endgroup$
    – user192680
    Nov 29, 2014 at 1:28
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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: $$\frac{5\times2^{3-t}\ln(2)}{(1+2^{-t})^2}=\frac{5\times2^{3-t}\ln(2)}{1+2\times2^{-t}+2^{-2t}}\times \frac{2^{2t}}{2^{2t}}=\frac{5\times2^{3+t}\ln(2)}{1+2\times2^{t}+2^{2t}}=\frac{5\times2^{3+t}\ln(2)}{(1+2^{t})^2}$$

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  • $\begingroup$ So then I'm not wrong? That is also the answer I get. The back of my text book provides a different answer for this problem which is $5ln(2) * 2^{3-2t}\over (1+2^{-t})^{2}$ Any idea why? $\endgroup$
    – Nolohice
    Nov 29, 2014 at 0:51
  • $\begingroup$ I'm not sure where the $2t$ comes from though, if you check here this gets: wolframalpha.com/input/?i=40%2F%281%2B2%5E%28-t%29%29 $\endgroup$
    – Jay
    Nov 29, 2014 at 0:54
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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.

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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

$$g'(t) = \frac{40.\ln 2.2^{-t}}{(1 + 2^{-t})^2}$$

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  • $\begingroup$ The answer you provided is actually the answer I keep coming up with. According to my text book I'm wrong, & the answer should be $5ln(2) * 2^{3-2t}\over (1 + 2^{-t})^{2}$ Not really sure how I'd turn that into the answer we came up with. $\endgroup$
    – Nolohice
    Nov 29, 2014 at 0:36
  • $\begingroup$ The $-2t$ exponent makes no sense and is wrong. But note however that $5\cdot 2^3 = 5\cdot 8 = 40$, so that part ties out. $\endgroup$
    – Simon S
    Nov 29, 2014 at 1:08
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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!!!

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