Question

Find the derivatives of the function:$f\left(x\right)=\frac{3x\sqrt{2x+1}}{1-x}.$

Short answer

The derivative of $f\left(x\right)=\frac{3x\sqrt{2x+1}}{1-x}is\frac{-3x^2+9x+3}{\sqrt{2x+1}{(1-x)}^2}.$

Step-by-step solution

Given information 

The given expression is$f\left(x\right)=\frac{3x\sqrt{2x+1}}{1-x}.$

Simplification  

$$\begin{gathered} f\text{'}\left(x\right) \\ =\frac{d}{dx}\left(\frac{3x\sqrt{2x+1}}{1-x}\right) \\ =\frac{\left(1-x\right){\displaystyle \frac{d}{dx}}\left(3x\sqrt{2x+1}\right)-\left(3x\sqrt{2x+1}\right){\displaystyle \frac{d}{dx}}\left\{\left(1-x\right)\right\}}{{(1-x)}^2} \\ =\frac{(3\sqrt{2x-1}+3x\times {\displaystyle \frac{1}{2\sqrt{2x+1}}}\times 2)-\{\left(1-x\right)-3x\sqrt{2x+1}\times \left(-1\right)}{{\left(1-x\right)}^2} \\ =\frac{-3x^2+9x+3}{\sqrt{2x+1}{\left(1-x\right)}^2} \\ Hence,f\text{'}\left(x\right)=\frac{-3x^2+9x+3}{\sqrt{2x+1}{\left(1-x\right)}^2}. \end{gathered}$$