hi, can someone explain why the derivative of y=2^x + 2^-x
goes to dy/dx = 2^xln2 - 2^-xln2
I'm not sure where the - came from, i thought the derivative was just dy/dx = 2^xln2 + 2^-xln2
this is what i thought
#differential
lethal edge
heavy torrent
you found the derivative of 2^(-x) wrong
2^(-x) can be viewed as a composition of two functions
f(g(x)) = 2^(-x)
f(x) = 2^x
g(x) = -x
then we can apply the chain rule to find the derivative
d/dx [f(g(x))] = f'(g(x))g'(x)
you can also use quotient rule because 2^(-x) = 1/(2^x), so it is a quotient of f(x) = 1 and f(x) = 2^x
but i think its easier to apply chain rule
lethal edge
does this work?
lethal edge
sorry thats how i got taught to do the chain rule :p
deep crest
i mean
all good, nothing wrong here
just
i'll need to redo the work myself to verify it
in a different way
lethal edge
ok no problem
heavy torrent
yeah thats why i pinged yoavmal
deep crest
$\dv{2^{-x}}{2^x}$ is the part that confuses me
whole glacierBOT
yoavmal
deep crest
you can write it as $f(x)=\frac{1}{x}$ and $g(x)=2^x$
whole glacierBOT
yoavmal
deep crest
but you can also write it as $f(x)=2^x$ and $g(x)=-x$
whole glacierBOT
yoavmal
deep crest
which seems generally easier to me
lethal edge
yeh thats what cosmic said, i have just never done it like that so i wasnt sure
deep crest
try it this way
it's the same technique
at least
that'd be my recommendation since it's easier to work on
deep crest
**yoavmal**
moreso, to calculate this, i'd need to calculate the derivative of 2^-x and the derivative of 2^x
heavy torrent
not f'(x)g(x)g'(x), f'(g(x))g'(x)
lethal edge
ah i see
heavy torrent
f(x) = 2^x, g(x) = -x
f'(x) = ln(2)2^x, g'(x) = -1
f'(g(x)) = ln(2)2^(-x)
f'(g(x))g'(x) = ln(2)2^(-x) * (-1) = -ln(2)2^(-x)
lethal edge
heavy torrent
you're welcome