Undergrad How to Find the Derivative with Respect to ##r##?

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To find the derivative of a function with respect to r, the chain rule is applied. Given the derivative of N(r) with respect to log(r), the correct approach is to multiply this derivative by the derivative of log(r) with respect to r. This confirms that the relationship holds true even though N(r) is expressed as a function of log(r). The discussion emphasizes the validity of using the chain rule for this type of derivative calculation. Understanding this principle is essential for correctly computing derivatives involving logarithmic functions.
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I have a derivative of a function with respect to ##\log \left(r\right)##:

\begin{equation*}
\frac{dN\left(r\right)}{d \log\left(r\right)} = \frac{N}{\sqrt{2\pi} \log\left(\sigma\right)} \exp\left\{-\frac{\left[\log \left(r\right) - \log\left(r_M\right)\right]^2}{2 \left[\log\left(\sigma\right)\right]^2}\right\}
\end{equation*}

I need to know the derivative of this function with respect to ##r##, that is ##\frac{dN\left(r\right)}{dr}##, how shall I do this? I was told that I just need to multiply the function with the derivative of the logarithm, that is

\begin{equation*}
\frac{dN\left(r\right)}{dr} = \frac{dN\left(r\right)}{d \log\left(r\right)} \cdot \frac{d \log\left(r\right)}{dr}
\end{equation*}

Is this correct? Even though ##N\left(r\right)## is a function of ##\log\left(r\right)##?
 
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Yes this is correct, it is a well known theorem for the derivative of the composition of two functions (in your case the two functions are N(y) and y(r)=logr)), known as chain rule for derivatives.

https://en.wikipedia.org/wiki/Chain_rule
 
Yes, its correct, it is the chain rule in one variable.
 
Alright. Thank you for your help!
 
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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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