How do you differentiate #y=(x^2+4x+3)/sqrtx#?

Question

Ezekiel Alexander · Accepted Answer

#dy/dx=(3x^2+4x-3)/(2x^(3/2))#

What I'd prefer is to do is split up the fraction into its three parts by dividing each term by #sqrtx=x^(1/2)#:

#y=x^2/x^(1/2)+(4x)/x^(1/2)+3/x^(1/2)#

#y=x^(3/2)+4x^(1/2)+3x^(-1/2)#

Now instead of having to use the quotient rule, we can simply differentiate term-by-term using the power rule:

#dy/dx=3/2x^(1/2)+4*1/2x^(-1/2)+3*-1/2x^(-3/2)#

#dy/dx=(3x^(1/2))/2+2/x^(1/2)-3/(2x^(3/2))#

Finding a common denominator of #2x^(3/2)#:

#dy/dx=(3x^2+4x-3)/(2x^(3/2))#

Samantha Adkison · Answer

undefined To differentiate $ y = \frac{x^2 + 4x + 3}{\sqrt{x}} $, we can use the quotient rule of differentiation:

$$ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} $$

Where $ f(x) = x^2 + 4x + 3 $ and $ g(x) = \sqrt{x} $.

First, find the derivatives of $ f(x) $ and $ g(x) $:

$$ f'(x) = 2x + 4 $$
$$ g'(x) = \frac{1}{2\sqrt{x}} $$

Then, apply the quotient rule:

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

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