<s>[INST] Derivative of a Rational function $f(x)=\sqrt{2x-5\over3x+1}$ I'm trying to find the derivative of,
$$f(x)=\sqrt{2x-5\over3x+1}$$
I think I can change this into
$$f(x)= \left({2x-5 \over 3x+1}\right)^{1\over2}
\\ =[(2x-5)(3x+1)^{-1}]^{1 \over 2}$$
Am I not right??
From now, how do I continue??
 [/INST] You have right, from here you have to use this formula: $(u^n)'=n\cdot u^{n-1}\cdot u'$, where $u=(2x-5)(3x+1)^{-1}\Rightarrow u'=\frac{17}{(3x+1)^2}$ 

So you'll obtain: $\frac{1}{2} \cdot u^{-\frac{1}{2}} \cdot u'=\frac{1}{2}\cdot (2x-5)^{-\frac{1}{2}}(3x+1)^{\frac{1}{2}}\cdot\frac{17}{(3x+1)^2}=\frac{17}{2\sqrt{2x-5}(3x+1)^{\frac{3}{2}}}$

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