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In the midterm of probability in my university, they demand for us to calculate the derivative of $ h(x)=8\times\log_e{(\frac{x}{10})}$ and i find $ h(x)^`=\frac{8}{10x}$ but in the solution they write $ h(x)^`=\frac{8}{x}$ , i repeat the operation a lot of time but i can't find the same result, i know that this derivative was simple but i was really confused! how to find the same result and why we ignore 10 ? the second question, in the exponential low, we find $\lambda = \frac{1}{8}$ and in the correction of the teacher he write that the $E(x)=\frac{1}{8}$ not 8, i really want to know how!

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    $\begingroup$ 1) How do you find $\frac8{10x}$? A still wrong but less surprising answer would be $\frac8{x/10}$. 2) Hint: $\log(x/10)=\log(x)-\log(10)$, and the derivative of a constant is $0$. $\endgroup$ Apr 21 at 6:40

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Looks like you forgot the chain rule: $$\frac{d}{dx}8\ln\left(\frac{x}{10}\right)=\frac{8}{x/10}\cdot\frac{d}{dx}\frac{x}{10}=\frac{10\cdot8}{x}\cdot\frac{1}{10}=\frac{8}{x}.$$

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There are at least two ways to see this. User csch2 already showed you via the chain rule. Here's another way:

Since $\ln\left(\frac x{10}\right)=\ln(x)-\ln(10)$, and $\ln(10)$ is constant, the derivative of $\ln\left(\frac x{10}\right)$ is the same as that of $\ln(x)$.

In general, if $f(x)=\ln(ax)$, then $f'(x)=\frac1x$, and we can ignore the constant $a$ inside the logarithm. Geometrically this is because for the logarithm, stretching or compressing the graph in the $x$-direction is equivalent to moving it in the $y$-direction. (Which is related to the fact that moving the graph of the exponential function in the $x$-direction is equivalent to stretching it in the $y$-direction due to $e^{x-a}=e^{-a}e^x$.)

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