# How do you differentiate (x^2 + 8x + 3 )/ sqrtx using the quotient rule?

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{3 {x}^{2} - 3}{2 x \sqrt{x}}$

#### Explanation:

The given equation is
$f \left(x\right) = \frac{{x}^{2} + 8 x + 3}{\sqrt{x}}$

We use the formula for derivative of rational expression
d/dx(u/v)=(v⋅d/dx(u)-u⋅d/dx(v))/v^2

d/dx(f(x))=d/dx((x^2+8x+3)/sqrt(x))=(sqrt(x)⋅d/dx(x^2+8x+3)-(x^2+8x+3)⋅d/dx(sqrt(x)))/(sqrt(x))^2

d/dx(f(x))=(sqrt(x)⋅(2x+8)-(x^2+8x+3)⋅(1/(2sqrt(x))))/(sqrt(x))^2

d/dx(f(x))=(((2x)⋅(2x+8)-(x^2+8x+3))/(2sqrt(x)))/(sqrt(x))^2

d/dx(f(x))=((2x)⋅(2x+8)-(x^2+8x+3))/(2xsqrt(x))

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{4 {x}^{2} + 8 x - {x}^{2} - 8 x - 3}{2 x \sqrt{x}}$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{3 {x}^{2} - 3}{2 x \sqrt{x}}$

God bless....I hope the explanation is useful.