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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.

Arturo Magidin
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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. –  Sep 17 '11 at 23:06
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    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: So here you are supposed to the bit of algebra I showed in the first comment to turn this expression into a sum of (mulitples of) powers, and then take the derivative by taking the derivative of each summand and adding the results. If you mistyped and meant "product or quotient rule", instead of "sum or difference rule", you'll get to those soon. – Arturo Magidin Sep 17 '11 at 23:15
  • I get it now, I don't know what I was doing earlier. It makes sense to me now. Thank you. –  Sep 17 '11 at 23:21
  • @Arturo Actually one thing I am a little confused on, when I am differentiating and I have a negative exponent that doesn't no matter when I bring it down does it? An example is the $ 3x^-(1/2) differentiates to 3(1/2), 3(-1/2)? I am getting $3/2 x ^ (1/3) + 2x ^-(1/2) - ? $ –  Sep 17 '11 at 23:31
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    @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
  • That makes sense, thank you. –  Sep 17 '11 at 23:51

2 Answers2

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Dividing through, your expression equals $$ x^{3/2}+4x^{1/2}+3x^{-1/2}. $$ Now you can use the power rule.

yunone
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You can also exercise your quotient rule: $\left( \dfrac{f}{g}\right)^'=\dfrac{f'g-gf'}{g^2}$.

Hassane Kone
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