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Question $\sqrt{2x+1}$
$=\sqrt{2x+1}$
$=\sqrt{2x} + \sqrt{1}$

$=\dfrac{1}{2x^{1/2}}$

however the right answer is $\dfrac{1}{\sqrt{2x+1}}$

Can you please help me out?
this chapter name is (differentiationg rational power $x^{p/q}$)

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    Not only is $\sqrt{2x+1}$ not at all the same thing as $\sqrt{2x}+\sqrt{1}$, but you shouldn't write $\sqrt{2x} + \sqrt{1} = \dfrac{1}{2x^{1/2}}$ if you mean $\dfrac{d}{dx}\left(\sqrt{2x} + \sqrt{1}\right) = \dfrac{1}{2x^{1/2}}$. The symbol "$=$" means "equals". You should use it only when you mean "equals".2012-04-03
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    @MichaelHardy the question is only $\sqrt{2x+1}$2012-04-03
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    Sb: $\sqrt{2x + 1}$ *is NOT a question!!!*... It is an expression. If you are asked to take the derivative of this, that's another story.2012-04-03
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    aww sorry. Can you plz explain me more? thx2012-04-03
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    How did you get to study derivatives if you write $$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$?2012-04-04

3 Answers 3

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First, $\sqrt{2x+1}$ is not $\sqrt{2x}+\sqrt1$. Second, don't write "first expression = second expression" when what you really mean is "derivative of first expression = second expression", as you've done when you wrote $\sqrt{2x}+\sqrt1=1/2x^{1/2}$. Third, the derivative of $\sqrt{2x}$ isn't $1/(2x^{1/2})$.

Other than that, everything is fine....

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    thx, but I'm not understood yet, can u plz explain more?2012-04-03
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    Sure. What would you like me to explain?2012-04-03
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    I understand now $(2x+1)^1/2$
    1/2(2x+1)1/2-1
    $\dfrac{1}{\sqrt{2x+1}}$
    2012-04-04
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$\sqrt{2x+1} \neq \sqrt{2x}+\sqrt{1}$: Witness $3 = \sqrt{9} = \sqrt{2 \cdot 4 + 1} \neq \sqrt{2 \cdot 4} + \sqrt{1} = 2 \sqrt{2} + 1$.

The right way to solve this is to apply the chain rule.

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Let $y = \sqrt{2x+1}$.

The problem is to find $\dfrac{d}{dx} \sqrt{2x+1}$, i.e. to find $\dfrac{dy}{dx}$.

Let $u=2x+1$. Then $y=\sqrt{u}$.

Then we have $$ \frac{dy}{du} = \frac{1}{2\sqrt{u}},\qquad \text{and}\qquad \frac{du}{dx} = 2. $$

Therefore $$ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = \frac{1}{2\sqrt{u}}\cdot 2 = \frac{1}{\sqrt{u}} = \frac{1}{\sqrt{2x+1}}. $$

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    I would like to know where is that $$ \frac{dy}{du} = \frac{1}{2\sqrt{u}} $$ come from?2012-04-04
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    Can u explanin? @MichaelHardy2012-04-04
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    There are several ways to show that $\dfrac{d}{du}\sqrt{u}$ is $\dfrac{1}{\sqrt{u}}$. One of those is the definition of "derivative": $\dfrac{d}{du}\sqrt{u} = \lim\limits_{\Delta u\to 0}\dfrac{\sqrt{u+\Delta u}-\sqrt{u}}{\Delta u}$. Then ratioanlize the numerator and simplify, and finally take the limit.2012-04-04
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    Here's another way: Let $y=\sqrt{u}$. Then $y^2=u$. Differentiate both sides with respect to $u$ to get $2y\dfrac{dy}{du}= 1$ (the chain rule was used in that step). From that you get $\dfrac{dy}{du}=\dfrac{1}{2y}$ and then finally $\dfrac{dy}{du}=\dfrac{1}{2\sqrt{u}}$.2012-04-04