What is the derivative of $f(x)= 3^{-x}$?

Question

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?

Answer 1

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 $\frac 1{3^x} = \left(\frac 13\right)^x,$ which allows you to write $y = f(x) = \left(\frac 13\right)^x,$ 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 $\ln\left(\frac 13\right) = - \ln 3$ as well as the fact (which you had already used once) that $3^{-x} = \left(\frac 13\right)^x.$

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As an aside, I'd like to observe that the rule "if $y=\left(\frac {a}{b}\right)^x$ then $\dot{y}=\left(\frac {a}{b}\right)^x \ln\left(\frac {a}{b}\right),$" 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.

Answer 2

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

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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 $(\frac 13)^x\ln \frac 13$.

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

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

Answer 3

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

Answer 4

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)

Answer 5

No,

$$(a^x)'=a^x\ln a$$

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

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