- #1
Karol
- 1,380
- 22
I am new to this forum, i don't know if it's here i should post this simple question.
I have to find the peak of the function:
##\frac{x}{\sqrt{x^2+R^2}(x^2+R^2)}=\frac{x}{(x^2+R^2)^{3/2}}##
I differentiate:
##\left( \frac{x}{(x^2+R^2)^{3/2}} \right)'=\frac{(x^2+R^2)^{3/2}+x\left( \frac{3}{2}(x^2+R^2)^{3\2}\cdot 2x \right)}{(x^2+R^2)^3}##
Only the numerator:
##(x^2+R^2)^{1/2}\left[ (x^2+R^2)^3+3x^2 \right]##
When i equal it to 0 i get, as one possibility:
##(x^2+R^2)^3=-3x^2##
And it's not good, since x has a value.
I have to find the peak of the function:
##\frac{x}{\sqrt{x^2+R^2}(x^2+R^2)}=\frac{x}{(x^2+R^2)^{3/2}}##
I differentiate:
##\left( \frac{x}{(x^2+R^2)^{3/2}} \right)'=\frac{(x^2+R^2)^{3/2}+x\left( \frac{3}{2}(x^2+R^2)^{3\2}\cdot 2x \right)}{(x^2+R^2)^3}##
Only the numerator:
##(x^2+R^2)^{1/2}\left[ (x^2+R^2)^3+3x^2 \right]##
When i equal it to 0 i get, as one possibility:
##(x^2+R^2)^3=-3x^2##
And it's not good, since x has a value.