Question

What method of differentiation is necessary, and what is the derivative?

Answer 1

The given function is

$f(x)=In\frac{2x^3}{\sqrt[]{x}}$

We need to use the derivation of some standard functions.

Recall that

$\frac{d}{dx}\text{In u(x)}=\frac{1}{x}u^{\prime}.$

Derivation of rational functions with quotient rule.

$\frac{d}{dx}(\frac{u(x)}{v(x)})=\frac{vu^{\prime}-uv^{\prime}}{v^2}$

Differentiate the given function with respect to x, we get

$f^{\prime}(x)=\frac{d}{dx}(In\frac{2x^3}{\sqrt[]{x}})$

$\text{Use }\frac{d}{dx}\text{In u(x)}=\frac{1}{x}u^{\prime}.$

$f^{\prime}(x)=\frac{1}{\frac{2x^3}{\sqrt[]{x}}}\frac{d}{dx}(\frac{2x^3}{\sqrt[]{x}})$

$f^{\prime}(x)=\frac{\sqrt[]{x}}{2x^3}\frac{d}{dx}(\frac{2x^3}{\sqrt[]{x}})$

$\text{Use }\frac{d}{dx}(\frac{u(x)}{v(x)})=\frac{vu^{\prime}-uv^{\prime}}{v^2}.$

$f^{\prime}(x)=\frac{\sqrt[]{x}}{2x^3}\times\frac{\sqrt[]{x}(2\times3x^2)-2x^3(\frac{1}{2}x^{-\frac{1}{2}})}{(\sqrt[]{x})^2}$

$f^{\prime}(x)=\frac{\sqrt[]{x}}{2x^3}\times\frac{6x^2\sqrt[]{x}-\frac{x^3}{\sqrt[]{x}}}{x}$

$f^{\prime}(x)=\frac{\sqrt[]{x}}{2x^3}\times\frac{\frac{6x^2\sqrt[]{x}\times\sqrt[]{x}}{\sqrt[]{x}}-\frac{x^3}{\sqrt[]{x}}}{x}$

$f^{\prime}(x)=\frac{\sqrt[]{x}}{2x^3}\times\frac{\frac{6x^3-x^3}{\sqrt[]{x}}}{x}$

$f^{\prime}(x)=\frac{\sqrt[]{x}}{2x^3}\times\frac{5x^3^{}}{\sqrt[]{x}}\times\frac{1}{x}$

$f^{\prime}(x)=\frac{\sqrt[]{x}}{2x^3}\times\frac{5x^3^{}}{\sqrt[]{x}}\times\frac{1}{x}$

$f^{\prime}(x)=\frac{5}{2x}$

Hence the required differentiation is

$f^{\prime}(x)=\frac{5}{2x}$