A derivative equality for density+distribution funs of N(0,1)

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An equality for density+distribution funs of N(0,1)

Reading through a book, I met the following equality (##F## is cumulative distribuion function, ##f## is density function)
$$\frac{d}{d\sigma}F_{N(0,\sigma^2)}(x)=\frac{-x}{\sigma}f_{N(0,\sigma^2)}(x)$$
which was given without any futher explanations (assumed obvious, I guess) but I have a hard time figuring out why it holds.

Could anyone, please, provide a hint on how to prove it?

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Mentor
What happens if you derive both sides with respect to x? You can exchange the order of the derivatives.
What happens for x -> minus infinity?

Reading through a book, I met the following equality (##F## is cumulative distribuion function, ##f## is density function)
$$\frac{d}{d\sigma}F_{N(0,\sigma^2)}(x)=\frac{-x}{\sigma}f_{N(0,\sigma^2)}(x)$$
which was given without any futher explanations (assumed obvious, I guess) but I have a hard time figuring out why it holds.

Could anyone, please, provide a hint on how to prove it?

It looks like you are dealing with a normal distribution. dF/dσ = (dF/dx)dx/dσ, where dx/dσ=-x/σ

Mentor
It looks like you are dealing with a normal distribution. dF/dσ = (dF/dx)dx/dσ, where dx/dσ=-x/σ
Why don't we have to treat σ and x as two independent variables here?

I may have mislead you. However I looked at it using brute force and the result comes out. It requires integration by parts in the middle of the derivation, since you will get a terms looking similar to F but with x2 in the integrand.

Mentor
With the method I described, you don't need any integrals, and you get the right result with two easy derivatives.