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As I'm trying to improve my maths I've now reached the subject of differentiation of more general exponential functions.

I've already studied the so called "chain rule".

Now, my book tells me:

$$y=a^x \rightarrow y^{\prime}=a^x\ln(a)$$

Then, there's the following function:

$$f(x) = 10^{-x}$$

Applying the above rule, this would lead me to believe that

$$f^{\prime}(x) = 10^{-x} \ln(10)$$

However, the book tells me it's

$$f^{\prime}(x) = -10^{-x} \ln(10)$$

Why's that? They say they also used the chain rule, but if I use the chain rule I get something totally different:

$$f(x) = 10^{-x}$$ $$f^{\prime}(x) = -x(10)^{-x-1}\cdot(-1*1)$$ $$f^{\prime}(x) = x(10)^{-x-1}$$

What am I doing wrong in all these cases?

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I don't think this should be tagged algebra-precalculus. – Aryabhata Oct 10 '10 at 15:05
up vote 6 down vote accepted

Let $N: \mathbb{R} \to \mathbb{R}$ be the negation function, i.e. $N(x) = -x$ and let $g(x) = 10^x$. Then it should be clear that your $f = g \circ N$.

$f'(x) = (g \circ N)'(x) = g'(N(x))N'(x) = \log 10 \cdot 10^{-x} \cdot (-1)$

PS: Learn to use $\LaTeX$ instead of code. It takes up less space and looks prettier.

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Thanks. Does this mean that the rule I mentioned is formulated incorrectly though? – Tom Oct 10 '10 at 9:36
In your first answer you tried to apply a formula for the derivative that doesn't apply directly in this particular case. As for your attempt at using the chain rule... I'm not sure. It looks like you're trying to treat $10^x$ as a polynomial. All I can say is that it doesn't work that way. – kahen Oct 10 '10 at 9:42
You applied the rule $(x^n)'=n\cdot x^{n-1}$ to the function $10^x$. That's not correct. It is also not the chain rule. – Rasmus Oct 10 '10 at 9:44
Alright, I'm still a bit confused as to when to use that rule but I think I will manage now. – Tom Oct 10 '10 at 9:46

The biggest problem here is that you seem to be confusing the chain rule with the power rule.

The power rule is as follows: Suppose you have some real number $n$. Then the function $f(x) =x^n$ has derivative $f'(x)=n\cdot x^{n-1}$. Notice that $n$ is fixed, but $x$ is variable.

The chain rule is as follows: Given functions $f$ and $g$, the function $h(x)=(f\circ g)(x) = f(g(x))$ has derivative $h'(x)= (f'\circ g)(x)\cdot g'(x)= f'(g(x))\cdot g'(x)$.

Now, the function you have is $f(x)=10^{-x}$. In this case, your exponent is variable, but $10$, obviously, is fixed. So the power rule does not apply; that is, we cannot simply bring down the exponent, multiply, and subtract one from the exponent, which is part of what you did in your work.

Also, given an exponential function $g(x)=a^x$, the derivative is $g'(x)=a^x\cdot\ln{(a)}$. So why does this rule not apply directly to the function $f(x)=10^{-x}$? Well, notice that the exponent in your $f$ is actually itself a function, namely, $-x$. The exponent in the rule from your book is merely $x$. That's why we have to apply the chain rule; we technically have a function "inside" another function. From here, follow kahen's solution above.

I again want to emphasize that the power rule is not applied to your $f$ because $f$ is exponential: it has a fixed base of $10$ and a variable exponent $-x$ (which is also a function).

Lastly, I should note that the power rule and chain rule as I have written them are neither rigorous nor precise, but I sacrificed details for readability.

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Great, thanks a lot. I am not sure whether to change the accepted answer to this one as it was more helpful. Is that common practice? – Tom Oct 11 '10 at 16:04
I have no idea. I've only been posting here about a week. – Bey Oct 11 '10 at 21:01

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