Question

Simplify the expression for $f\left(x\right)$ and then find $f^{\prime }\left(x\right)$, if
$f\left(x\right)(\frac{2\sqrt{x}+3\sqrt[4]{4x}}{\sqrt{16x}+12\sqrt[4]{4x}+9}-\frac{\sqrt[4]{4x}-3}{2\sqrt[4]{4x}+3})({2.3}^{lo{g}_{0.2x}}+3)$

Answer 1

$f\left(x\right)=(\frac{2\sqrt{x}+3\sqrt[4]{4x}}{\sqrt{16x}+12\sqrt[4]{4x}+9}-\frac{\sqrt[4]{4x}-3}{2\sqrt[4]{4x}+3})({2.3}^{lo{g}_{0.2x}}+3)$

$\sqrt{16x}+12\sqrt[4]{4x}+9=(2\sqrt[4]{4x}{)}^2+2\ast 3\ast 2\sqrt[4]{4x}+3^2=(2\sqrt[4]{4x}+3{)}^2$

Using this, we get

$\frac{2\sqrt{x}+3\sqrt[4]{4x}}{\sqrt{16x}+12\sqrt[4]{4x}+9}-\frac{\sqrt[4]{4x}-3}{2\sqrt[4]{4x}+3}$

$=\frac{\sqrt[4]{4x}(2\sqrt[4]{4x}+3)}{\sqrt{16x}+12\sqrt[4]{4x}+9}-\frac{\sqrt[4]{4x}-3}{2\sqrt[4]{4x}+3}$

$=\frac{\sqrt[4]{4x}(2\sqrt[4]{4x}+3)}{(2\sqrt[4]{4x}+3{)}^2}-\frac{\sqrt[4]{4x}-3}{2\sqrt[4]{4x}+3}$

$=\frac{\sqrt[4]{4x}}{2\sqrt[4]{4x}+3}-\frac{\sqrt[4]{4x}-3}{2\sqrt[4]{4x}+3}$

$=\frac{3}{2\sqrt[4]{4x}+3}$

$({2.3}^{lo{g}_{81}x}+3)=\left({2.3}^{\frac{1}{4}{\log }_{3}x}\right)=(2\sqrt[4]{4x}+3)$

$f\left(x\right)$

$=(\frac{2\sqrt{x}+3\sqrt[4]{4x}}{\sqrt{16x}+12\sqrt[4]{4x}+9}-\frac{\sqrt[4]{4x}-3}{2\sqrt[4]{4x}+3})({2.3}^{lo{g}_{0.2x}}+3)$

$=\frac{3}{2\sqrt[4]{4x}+3}(2.\sqrt[4]{4x}+3)$

$=3$

$f^{\prime }\left(x\right)=0$