<s>[INST] 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
 [/INST] 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 } \\$
$$ f'\quad =2\cos { u } \frac { du }{ dx } =\frac { 2 }{ \sqrt { x^{ 2 }+1 }  } \frac { du }{ dx } =\frac { 2 }{ \sqrt { x^{ 2 }+1 }  } \frac { 1 }{ x^{ 2 }+1 } $$
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