How do you derive #y = (x^2+8x+3)/x^(1/2)# using the quotient rule?

Question

Isabella Ackerman · Accepted Answer

#(3x^2 + 8x - 3)/(2x^(3/2))# #=[(x^2 + 8x + 3)'(x^(1/2)) - (x^(1/2))'(x^2 +8x + 3)]/(x^(1/2))^2# #=[x^(1/2)(2x+8) - 1/2x^(-1/2)(x^2+8x+3)]/x# #=[2x^(3/2) + 8x^(1/2) - 1/2x^(3/2) - 4x^(1/2) - 3/2x^(-1/2)]/x# #=[3/2x^(3/2) + 4x^(1/2)-3/2x^(-1/2)]/x# #=[x^(-1/2)[3/2x^2 + 4x -3/2]]/x# #=[(3x^2 + 8x + 3)/2]/x^(3/2)# #=(3x^2 + 8x + 3)/(2x^(3/2))#

Isabella Ackerman · Answer

undefined To derive the function $ y = \frac{x^2 + 8x + 3}{\sqrt{x}} $ using the quotient rule:

Let $ u(x) = x^2 + 8x + 3 $ and $ v(x) = \sqrt{x} $.

Find the derivatives:
- $ u'(x) = 2x + 8 $
- $ v'(x) = \frac{1}{2\sqrt{x}} $ (using the power rule for differentiation)

Apply the quotient rule:
$$ y' = \frac{u'v - uv'}{v^2} $$

$$ y' = \frac{(2x + 8) \cdot \sqrt{x} - (x^2 + 8x + 3) \cdot \frac{1}{2\sqrt{x}}}{(\sqrt{x})^2} $$

$$ y' = \frac{(2x + 8)\sqrt{x} - \frac{x^2 + 8x + 3}{2\sqrt{x}}}{x} $$

$$ y' = \frac{(2x + 8)\sqrt{x} - \frac{x^2}{2\sqrt{x}} - \frac{8x}{2\sqrt{x}} - \frac{3}{2\sqrt{x}}}{x} $$

$$ y' = \frac{(2x + 8)\sqrt{x} - \frac{x^2}{2x^{1/2}} - \frac{8x}{2x^{1/2}} - \frac{3}{2x^{1/2}}}{x} $$

$$ y' = \frac{(2x + 8)\sqrt{x} - \frac{x^{3/2}}{2} - 4x^{1/2} - \frac{3}{2x^{1/2}}}{x} $$

$$ y' = \frac{(2x + 8)\sqrt{x} - \frac{x^{3/2}}{2} - 4\sqrt{x} - \frac{3}{2\sqrt{x}}}{x} $$

$$ y' = \frac{2x\sqrt{x} + 8\sqrt{x} - \frac{x^{3/2}}{2} - 4\sqrt{x} - \frac{3}{2\sqrt{x}}}{x} $$

$$ y' = \frac{2x\sqrt{x} + 4\sqrt{x} - \frac{x^{3/2}}{2} - \frac{3}{2\sqrt{x}}}{x} $$