# Find the derivative respect to $x$?

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

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

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

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". –  Michael Hardy Apr 3 '12 at 12:44
@MichaelHardy the question is only $\sqrt{2x+1}$ –  Sb Sangpi Apr 3 '12 at 12:53
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. –  The Chaz 2.0 Apr 3 '12 at 12:58
aww sorry. Can you plz explain me more? thx –  Sb Sangpi Apr 3 '12 at 13:03
How did you get to study derivatives if you write $$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$? –  Pedro Tamaroff Apr 4 '12 at 1:17

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? –  Sb Sangpi Apr 3 '12 at 12:53
Sure. What would you like me to explain? –  Gerry Myerson Apr 3 '12 at 13:16
I understand now $(2x+1)^1/2$ <br> 1/2(2x+1)1/2-1 <br>$\dfrac{1}{\sqrt{2x+1}}$ –  Sb Sangpi Apr 4 '12 at 1:31

$\sqrt{2x+1} \neq \sqrt{2x}+\sqrt{1}$: 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? –  Sb Sangpi Apr 4 '12 at 2:57
Can u explanin? @MichaelHardy –  Sb Sangpi Apr 4 '12 at 5:10
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. –  Michael Hardy Apr 4 '12 at 19:23
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 $\dfrac{dy}{du}=\dfrac{1}{2\sqrt{u}}$. –  Michael Hardy Apr 4 '12 at 19:26
thank you so much! –  Sb Sangpi Apr 4 '12 at 23:45