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Today in class we learned that for exponential functions $f(x) = b^x$ and their derivatives $f'(x)$, the ratio is always constant for any $x$. For example for $f(x) = 2^x$ and its derivative $f'(x) = 2^x \cdot \ln 2$

$$\begin{array}{c | c | c | c} x & f(x) & f'(x) & \frac{f'(x)}{f(x)}\\ \hline -1 & \frac{1}{2} & 0.346 & 0.693 \\ 0 & 1 & 0.693 & 0.693 \\ 1 & 2 & 1.38 & 0.693\\ 2 & 4 & 2.77 &0.693& \end{array}$$

So as you can see, the ratio is the same and this is true for all functions of the form $b^x$ and its derivative. So my question is, why is the ratio always constant? Is there some proof or logic behind it or is just like that? Furthermore, what's the use of knowing this?

EDIT:

It seems that I have missed the fairly simple

$$\require{cancel}\frac{f^\prime(x)}{f(x)}=\frac{\cancel{{b^x}}\ln\,b}{\cancel{{‌​b^x}}}=\ln\,b$$

But what's the use knowing and learning this? Will this reduce a step in the future or help solve a much harder problem more easily?

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It's not entirely clear to me if you're asking for a proof or for an intuitive reason why this happens. Also note that $\ln (2)=0.6931471806\ldots$ –  Git Gud Apr 9 '13 at 8:27
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$$\require{cancel}\frac{f^\prime(x)}{f(x)}=\frac{\cancel{{b^x}}\ln\,b}{\cancel{{‌​b^x}}}=\ln\,b$$ –  J. M. Apr 9 '13 at 8:28
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Note that the left hand side could be read as $[\ln f(x)]'$, so that $[\ln a^x]' = [x \ln a]' = \ln a$ is constant. –  TMM Apr 9 '13 at 8:36
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2 Answers 2

up vote 2 down vote accepted

$$f=a^x$$ $$f'=\ln a \times a^x$$ So, $$f'=\ln a \times f$$ So, $$\dfrac{f'}{f}=\ln a= \text{constant}=0.693\Big|_{a=2}$$

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What's the use of knowing this? Or in other words, why am I learning it? Is it going to make a future problem much easier to solve, what's its significance? –  gekkostate Apr 9 '13 at 8:32
    
There's no significance, if you have function as yours. And you get to calculate single differentiation of $\dfrac{f'}{f}$, its just $0$, you don't need to think hard for this. –  Inceptio Apr 9 '13 at 8:33
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@gekko, for now, treat it as a nice property of the exponential function. Later, when you get to solving differential equations, this fact might prove useful... –  J. M. Apr 9 '13 at 8:34
    
Yes , DE of a exponential function is $f'=Kf$. –  Mr.ØØ7 Apr 9 '13 at 8:36
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@Inceptio The only reason I'm thinking "hard" about this is because we spent about 75 mins learning this and I had no idea why and teacher didn't specify why this important or useful but I know why and I know it's significance - helps with differential equations. –  gekkostate Apr 9 '13 at 8:47
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$f'/f$ is called the Logarithmic Derivative of $f$.

By the Chain Rule $$ \frac{f'}{f}=\frac{\mathrm{d}}{\mathrm{d}x}\log(f(x)) $$ Since $\log(b^x)=x\log(b)$ and $\frac{\mathrm{d}}{\mathrm{d}x}x\log(b)=\log(b)$, we get that $f'/f$ is constant.

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