# Thread: Solve using the quotient rule

1. ## Solve using the quotient rule

Differentiate square root of x divided by x^2+3

The work i've done:

1/2x^(-1/2) (x^2+3) - 2x (x^ 1/2) / (x^2+3)^2

= 1/2x^(3/2) + 3/2x^(-1/2) - 2x^(3/2)

= -3/2x^(3/2) + 3/2x^(-1/2) / (x^2+3)^2

After that it's all blurry, not sure what to do....

thanks for the help in advance.

2. Originally Posted by agent2421

Differentiate square root of x divided by x^2+3

The work i've done:

1/2x^(-1/2) (x^2+3) - 2x (x^ 1/2) / (x^2+3)^2

= 1/2x^(3/2) + 3/2x^(-1/2) - 2x^(3/2)

= -3/2x^(3/2) + 3/2x^(-1/2) / (x^2+3)^2

After that it's all blurry, not sure what to do....

thanks for the help in advance.
$\dfrac{d}{dx} \left(\dfrac{\sqrt{x}}{x^2+3}\right)$

$\dfrac{(x^2+3) \cdot \frac{1}{2\sqrt{x}} - \sqrt{x} \cdot 2x}{(x^2+3)^2}$

$\dfrac{(x^2+3) - 2x \cdot 2x}{2\sqrt{x}(x^2+3)^2}$

$\dfrac{3(1 - x^2)}{2\sqrt{x}(x^2+3)^2}$

3. thanks... can you explain how you did the 3rd part though? When you brought 2 square root of x to the bottom... and multiplied the top by 2 x... i didn't get that part fully... also the last step where does 3 (1-x^2) come from?

4. Skeeter multiplied the numerator and denominator by $2\sqrt{x}$ to clear that fraction in the numerator hence $2\sqrt{x}$ in the denominator goes straight in.

The numerator is $\left(2\sqrt{x} \times (x^2+3)\right) \cdot \dfrac{1}{2\sqrt{x}} - \left(2\sqrt{x} \cdot \sqrt{x} \cdot 2x \right)$

$= (x^2+3) - 2x \cdot 2x = (x^2+3) - 4x^2 = 3-3x^2 = 3(1-x^2)$

5. 2nd to 3rd step ...

$\dfrac{2\sqrt{x}}{2\sqrt{x}} \cdot \dfrac{(x^2+3) \cdot \frac{1}{2\sqrt{x}} - \sqrt{x} \cdot 2x}{(x^2+3)^2}$

$\dfrac{(x^2+3) - 2x \cdot 2x}{2\sqrt{x}(x^2+3)^2}$

3rd to last step ... combine like terms and factor out the 3

$\dfrac{3(1 - x^2)}{2\sqrt{x}(x^2+3)^2}$