1st derivative of $\frac{2x}{\sqrt{x^2 + 1}}$ Another simple question that I can't work out today, yet I would work it out two weeks ago!
I need to find the 1st derivative of $$\frac{2x}{\sqrt{x^2 + 1}}$$.
So I use the Quotient rule and I get: $$\frac{(x^2 + 1)^.5 (2) - (2x)(0.5x^2 + 0.5)^-5}{x^2+1}$$
Am I heading in the correct direction and do I just need to multiply and try to get rid of the exponents somehow?
Thanks
 A: The numerator is not quite right. We want $(x^2+1)^{1/2}(2)-2x g'(x)$ where $g(x)=(x^2+1)^{1/2}$. That derivative will be $(2x)(1/2)(x^2+1)^{-1/2}$. 
The denominator is right. Now you can if it is useful do some manipulation ("simplification"). 
Remark: For square roots, I think you will find $\sqrt{x^2+1}$ easier and safer to work with than the exponent notation. For other roots, the advantage goes over to the exponent notation. 
A: $$\left(\frac{2x}{\sqrt{x^2 + 1}}\right)'=\frac{(2x)'(\sqrt{x^2 + 1})-(\sqrt{x^2 + 1})'2x}{(\sqrt{x^2 + 1})^2}.$$
A: I'm not sure if I read your notation correctly, but it looks like there's a mistake in the numerator:
$$f(x)=2x,\quad f^{\prime}(x)=2,\quad g(x)=\sqrt{x^2+1},\quad g^{\prime}(x)=
\frac{2x}{2\sqrt{x^2+1}}$$
And the numerator should then be $f^{\prime}(x)g(x)-f(x)g^{\prime}(x)$.
A: Without messing up with nominator and denominators:
$x=\tan { u } $ then draw right triangle, see
$\\ \frac { 1 }{ \sqrt { x^{ 2 }+1 }  } =\cos { u } \\$
$\\ \frac { x }{ \sqrt { x^{ 2 }+1 }  } =\sin { u } \\$
$ \frac { du }{ dx } =\frac { 1 }{ x^{ 2 }+1 } \\$
$ f=\frac { 2x }{ \sqrt { x^{ 2 }+1 }  } =2\sin { u } \\$
$$ f'\quad =2\cos { u } \frac { du }{ dx } =\frac { 2 }{ \sqrt { x^{ 2 }+1 }  } \frac { du }{ dx } =\frac { 2 }{ \sqrt { x^{ 2 }+1 }  } \frac { 1 }{ x^{ 2 }+1 } $$
