# Differentiating square root fractions

I am suppose to differentiate $(x^2 +4x +3)/ \sqrt{x}$ I know that a square root is equal to $x^{1/2}$ but I still am not able to properly differentiate this problem. I ended up with $(2x+4)/ (1/2)x^{1/2})$ which I know is wrong.

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You seem to be trying to differentiate the numerator and the denominator separately; that does not work. You have to either use the Quotient Rule or, if you don't know it yet, first simplify the expression using algebra:$$\frac{x^2+4x+3}{\sqrt{x}} = \frac{x^2+4x+3}{x^{1/2}} = \frac{x^2}{x^{1/2}}+\frac{4x}{x^{1/2}} + \frac{3}{x^{1/2}} = x^{3/2}+4x^{1/2}+3x^{-1/2}$$and then differentiate. –  Arturo Magidin Sep 17 '11 at 21:05
Hint: $\frac{a+b+c}{d}=\frac{a}{d}+\frac{b}{d}+\frac{c}{d}$. –  André Nicolas Sep 17 '11 at 21:07
I am suppose to be learning the sum, difference and power rule. I have not yet encountered the sum or difference rule and I have no idea how to use them. –  user138246 Sep 17 '11 at 23:06
The "sum rule" just says that $\frac{d}{dx}(f+g) = \frac{d}{dx}f + \frac{d}{dx}g$ (the derivative of a sum is the sum of the derivatives), and the "difference rule" just says that $\frac{d}{dx}(f-g) = \frac{d}{dx}f - \frac{d}{dx}g$, the derivative of the difference is the difference of the derivatives. In fact, you've used them alread, when you were trying to take the derivative of a polynomial by doing it for each term and then adding. This works for sums/difference, but not for products or quotients. –  Arturo Magidin Sep 17 '11 at 23:12
@Jordan: The power rule works for any exponent, so long as the base is just the variable by itself, and the exponent is constant. $\displaystyle \frac{d}{dx}x^n = nx^{n-1}$. So for $x^{-1/2}$, you get $$\frac{d}{dx}x^{-1/2} = -\frac{1}{2}x^{-\frac{1}{2}-1} = -\frac{1}{2}x^{-\frac{3}{2}}.$$Just be careful with the subtraction in the exponent. For $3x^{-1/2}$, you have: $$\frac{d}{dx}\left(3x^{-1/2}\right) = 3\frac{d}{dx}x^{-1/2} = 3\left(-\frac{1}{2}\right)x^{-\frac{3}{2}}= -\frac{3}{2}x^{-3/2}.$$The fact that the coefficient equals the exponent here is coincidence, don't read anything into it. –  Arturo Magidin Sep 17 '11 at 23:46

Dividing through, your expression equals $$x^{3/2}+4x^{1/2}+3x^{-1/2}.$$ Now you can use the power rule.
You can also exercise your quotient rule: $\left( \dfrac{f}{g}\right)^'=\dfrac{f'g-gf'}{g^2}$.