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I have recently started learning derivatives and I came across this simple looking problem. What is the derivative of $f(x)=3^{-x}$. When I use different "formlas" I always get a different result. For eg. first I tried using the formula $$y=a^x$$
$$dot {y}=a^xlna$$
which results in $dot {y}=3^{-x}ln3$.

After that I tried rewriting it as $f(x)=frac {1}{3^x}$ and applied the same formula which resulted in $dot {y}=frac {1}{3^xln3}$.

I also tried using the formula$$y=(frac {a}{b})^x$$
$$dot {y}=(frac {a}{b})^xln(frac {a}{b})$$
which results in $dot {y}=(frac {1}{3})^xln(frac {1}{3})$

I know the answer should be $dot {y}=-3^{-x}ln3$, but I can't seem to get to that result using the formulas. Does that mean the formulas are not correct? And how can I get to the correct result?

The thing about formulaic solutions such as
"if $y=a^x$, then $dot y = a^x ln a$,"
is that they have to be applied exactly as written.
You may substitute any reasonable constant number for $a$ (where "reasonable" in this case means "positive," so that the log will be defined),
but that is all you can do with this formula.

So, given $y = 3^{-x},$ your first attempt, substituting $a = 3,$
fails because then $a^x = 3^x,$ not $3^{-x}$ as required.
Indeed this substitution tells us that if $y = 3^x$ then
$dot y = 3^x ln 3,$ but that wasn't the problem you wanted to solve.

Your second attempt, writing $y = frac 1{3^x},$
does not match the rule at all, since the rule requires
$y = a^x$ and not $y = frac 1{a^x}.$
The application of the rule after that point is simply false.

The second attempt was close, however.
You just needed to see that $frac 1{3^x} = left(frac 13right)^x,$
which allows you to write $y = f(x) = left(frac 13right)^x,$
at which point you would have a very good chance to apply the formula for $y = a^x$ correctly.

As already pointed out in another answer, the third attempt, with the substitution $frac ab = frac 13,$
was completely successful.
You just needed to finish simplifying the result, using the fact that
$lnleft(frac 13right) = - ln 3$
as well as the fact (which you had already used once)
that $3^{-x} = left(frac 13right)^x.$

As an aside, I'd like to observe
that the rule "if $y=left(frac {a}{b}right)^x$ then
$dot{y}=left(frac {a}{b}right)^x lnleft(frac {a}{b}right),$"
while it is a perfectly good rule, is practically the same rule as
"if $y=a^x$, then $dot y = a^x ln a$."
That is, the only difference in the rules is that in one rule we have a constant written $a$ and in the other we have a constant written $frac ab.$
One of the things you can do with rules like this (while still using them "exactly as written") is to replace a constant written one way with a constant written a different way. You just need to be sure you change the constant everywhere in the formula without missing any place where it occurs.

Just use the chain rule $f(x) = 3^x$ and $g(x) = -x$ so $h(x) = 3^{-x} = f(g(x))$ and $h'(x) = f'(g(x))*g'(x) = ln (3) 3^{-x}*(-1) = -ln (3) 3^{-x}$.

...

Your first method doesn't work because $3^x ne 3^{-x}$ and it doesn't apply.

Your second method was a perfect success.

You got $(frac 13)^xln frac 13$.

You state the answer should be $-3^{-x}ln 3$. Which is the the same thing as what you got.

$-3^{-x}ln 3 = (frac 13)^x ln frac 13$.

You can apply the quocient rule:

$frac{u'v - uv'}{v^2}$

Where $f(x) = frac{1}{3^x}$

So $u = 1$ and $v = 3^x$

$frac{(1)'3^x - (1)*(3^x)'}{3^{2x}}$

You have:

$-frac{3^{x}ln(3)}{3^{2x}}$

Finally:

$-frac{ln(3)}{3^x}$ or $-3^{-x}ln(3)$

Option:

Set $e^a=3$, where $a= log 3 >1$, real.

$f(x)=e^{-ax}$;

$f'(x)=(-a)e^{-ax}= (- log 3)3^{-x}.$
(Chain rule)

No,

$$(a^x)'=a^xln a$$

does not imply that

$$(3^{-x})'=3^{-x}ln 3$$ because there is a minus sign.

You can deal with it as

• $3^{-x}=(3^{-1})^x$, giving $3^{-x}ln(3^{-1})=-3^{-x}ln 3$ (because $a=3^{-1}$), or




• $dfrac1{3^x}$, giving $-dfrac{3^xln 3}{(3^x)^2}=-dfrac{ln 3}{3^x}$ (by the derivative of the inverse of a function), or




• $3^{(-x)}$ giving $3^{-x}(-x)'ln3=-3^{-x}ln3$ (by the chain rule).