# Differentiate the function: y= \frac {8x^2+8x+2}{\sqrt x}

## Question:

Differentiate the function: {eq}y= \frac {8x^2+8x+2}{\sqrt x} {/eq}

## Differentiating Function Using Quotient Rule

Given a function that is a quotient of two functions, we use the quotient rule from Calculus to find the first derivative of the function. The quotient rule for the first derivative is as follows: {eq}\displaystyle \left( \frac uv \right)' = \frac {vu'-v'u }{v^2} {/eq}

## Answer and Explanation:

Given the function

{eq}\displaystyle y = \frac {8x^2+8x+2}{\sqrt {x}} \qquad (1) {/eq}

we use the quotient rule described in the Content section to obtain

{eq}\displaystyle y' = \frac {dy}{dx} = \frac {\sqrt {x} (16x+8) - (8x^2+8x+2)(1/2\sqrt {x})}{x} \qquad (2) {/eq}

Multiplying the numerator and denominator of (2) above by the square root of x gives us

{eq}\displaystyle y' = \frac {dy}{dx} = \frac {x(16x+8) - (4x^2+4x+1)}{x^{3/2}} = \frac {12x^2+4x-1}{x^{3/2}}. {/eq}

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