Question

$f\left(x\right)=\sqrt{\frac{2x+1}{x^3-3x^2+2x}}$.

Answer 1

$f\left(x\right)=\sqrt{\frac{2x+1}{x^2-3x^{+}2x}=\sqrt{\frac{2x+1}{x(x-2)(x-1)}}}$

$\left[f\right(x){]}^2=\frac{}{2}x+1x(x-2)(x-1)=\frac{A}{x}+\frac{B}{(x-2)}+\frac{C}{(x-1)}$

simplifying $2x+1=A(x-2)(x-1)+B(x-1)x+C(x-2)x$

$2x+1=At{x}^2(A+B+C)+x(-3A-C-2C)+A(2A+0+0)$

comparing coefficient

$2A=1\Rightarrow A=1/2A+B+C=0\Rightarrow B+C=-1/2$

$2=-.A-B-2C\Rightarrow 2=\frac{-3}{2}-B-2C\Rightarrow B+2C=-\frac{3}{2}-2=-\frac{7}{2}$

$\therefore B+C=-1/2,B+2C=-7/2$

solving $\Rightarrow C=-3,B=5/2,A=1/2$

$\therefore \left[f\right(x){]}^2=\frac{1}{2x}+\frac{5}{2(x-2)}+\frac{3}{(x-1)}$

differentiating w.r.t ${}^{\prime }{x}^{\prime }$

$2f\left(x\right),f^{\prime }\left(x\right)=\frac{-1}{2x^2}+\frac{-5}{2(x-2{)}^2}+\frac{3}{(x-1{)}^2}$

$f^{\prime }\left(x\right)=\frac{1}{2\sqrt{\frac{2x+1}{x(x-2)(x-1)}}}=[\frac{1}{2x^2}-\frac{5}{2(x-2{)}^2+\frac{3}{(x-1{)}^2}}]$