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

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

Derivation of rational functions with quotient rule.

`(d)/(dx)((u(x))/(v(x)))=(vu^(\prime)-uv^(\prime))/(v^2)`

Differentiate the given function with respect to x, we get

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

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

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

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

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

`f^(\prime)(x)=\frac{\sqrt[]{x}}{2x^3}*\frac{\sqrt[]{x}(2*3x^2)-2x^3((1)/(2)x^{-(1)/(2)})}{(\sqrt[]{x})^2}`

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

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

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

`f^(\prime)(x)=\frac{\sqrt[]{x}}{2x^3}*\frac{5x^3^{}}{\sqrt[]{x}}*(1)/(x)`

`f^(\prime)(x)=\frac{\sqrt[]{x}}{2x^3}*\frac{5x^3^{}}{\sqrt[]{x}}*(1)/(x)`

`f^(\prime)(x)=(5)/(2x)`

Hence the required differentiation is

`f^(\prime)(x)=(5)/(2x)`