What is the Derivative of a Fraction with a Square Root in the Denominator?

[Karol]

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.

 

[Mentallic]

$(x^2+R^2)^{1/2}\left[ (x^2+R^2)^3+3x^2 \right]$

You've factorized incorrectly here.

Also, your derivative is incorrect. Remember u'v - v'u, as opposed to +.

 

[Karol]

Thanks