# Find the derivative for the function: f(x)=\frac {2x^2-1}{2x^{1/2}+x}

## Question:

Find the derivative for the function: {eq}f(x)=\frac {2x^2-1}{2x^{1/2}+x} {/eq}

## Quotient Rule and Power Rule:

The quotient rule and power rule are two of the introductory differentiation rules we need to learn.

While the quotient rule deals with the derivative of a quotient of two functions, the power rule takes on the derivative of a power function.

Quotient Rule: {eq}\displaystyle \left(\frac{g(x)}{h(x)}\right)' = \frac{h(x)g'(x) -g(x)h'(x)}{(h(x))^2} {/eq}

Power Rule: {eq}\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}x^a=ax^{a-1} {/eq}

The function {eq}\displaystyle f(x)=\frac {2x^2-1}{2x^{\frac{1}{2}}+x} {/eq} is a quotient of the functions {eq}2x^2-1 {/eq} and {eq}2x^{\frac{1}{2}}+x {/eq} so we will calculate its derivative by implementing the quotient rule.

After utilizing the quotient rule, the power rule will also be employed as can be seen below:

{eq}\begin{align*} \displaystyle f(x)&=\frac {2x^2-1}{2x^{\frac{1}{2}}+x}\\ f'(x)&=\frac{(2x^{\frac{1}{2}}+x)\frac{\mathrm{d}}{\mathrm{d}x}(2x^2-1)-(2x^2-1)\frac{\mathrm{d}}{\mathrm{d}x}(2x^{\frac{1}{2}}+x)}{(2x^{\frac{1}{2}}+x)^2} && \left[\mathrm{ Quotient \ Rule}\right]\\ f'(x)&=\frac{(2x^{\frac{1}{2}}+x) (4x)-(2x^2-1)\left(\frac{1}{\sqrt{x}}+1\right)}{(2x^{\frac{1}{2}}+x)^2} && \left[\mathrm{ Power \ Rule}\right]\\ f'(x)&=\frac{ \frac{2x^2\sqrt{x}+6x^{2}+\sqrt{x}+1}{\sqrt{x}}}{(2x^{\frac{1}{2}}+x)^2} && \left[\mathrm{Expand \ and \ simplify \ terms \ in \ the \ numerator }\right]\\ f'(x)&=\frac{ 2x^2\sqrt{x}+6x^{2}+\sqrt{x}+1}{\sqrt{x}(2x^{\frac{1}{2}}+x)^2} \\ \end{align*} {/eq}

Therefore, {eq}\displaystyle f'(x)=\frac{ 2x^2\sqrt{x}+6x^{2}+\sqrt{x}+1}{\sqrt{x}(2x^{\frac{1}{2}}+x)^2} {/eq} is the derivative for the function {eq}\displaystyle f(x)=\frac {2x^2-1}{2x^{\frac{1}{2}}+x} {/eq}.