Q. 28 - Calculus

Question

Find the derivatives of the functions in Exercises 21–46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating.  

$f\left(x\right)=\frac{{\left(1+\sqrt{x}\right)}^2}{3x^2-4x+1}$

Step-by-Step Solution

Answer

The required answer is $\frac{1+\sqrt{x}}{\sqrt{x}(3x^2-4x+1)}-\frac{{\left(1+\sqrt{x}\right)}^2(6x-4)}{{\left(3x^2-4x+1\right)}^2}$

Step 1. Given Information

The given function is $f\left(x\right)=\frac{{\left(1+\sqrt{x}\right)}^2}{3x^2-4x+1}$

Step 2. Calculation

Differentiate both the sides with respect to x, we get,     

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\left({\displaystyle \frac{d}{dx}}{\left(1+\sqrt{x}\right)}^2\right)\left(3x^2-4x+1\right)-{\left(1+\sqrt{x}\right)}^2\left({\displaystyle \frac{d}{dx}}\left(3x^2-4x+1\right)\right)}{{\left(3x^2-4x+1\right)}^2} \\ =\frac{\left({\displaystyle 2\left(1+\sqrt{x}\right)\frac{d}{dx}\left(1+\sqrt{x}\right)}\right)\left(3x^2-4x+1\right)-{\left(1+\sqrt{x}\right)}^2\left({\displaystyle 3\left(2x\right)-4}\right)}{{\left(3x^2-4x+1\right)}^2} \\ =\frac{\left({\displaystyle 2\left(1+\sqrt{x}\right)\frac{1}{2}{x}^{-\frac{1}{2}}}\right)\left(3x^2-4x+1\right)-{\left(1+\sqrt{x}\right)}^2\left({\displaystyle 6x)-4}\right)}{{\left(3x^2-4x+1\right)}^2} \\ =\frac{\left({\displaystyle \frac{1+\sqrt{x}}{\sqrt{x}}}\right)\left(3x^2-4x+1\right)-{\left(1+\sqrt{x}\right)}^2\left({\displaystyle 6x)-4}\right)}{{\left(3x^2-4x+1\right)}^2} \\ =\frac{\left({\displaystyle 1+\sqrt{x}}\right)}{\sqrt{x}{\left(3x^2-4x+1\right)}^2}-\frac{{\left(1+\sqrt{x}\right)}^2\left({\displaystyle 6x)-4}\right)}{{\left(3x^2-4x+1\right)}^2} \end{gathered}$$