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 $x^{p/q}$)
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 $x^{p/q}$)
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....
$\sqrt{2x+1} \neq \sqrt{2x}+\sqrt{1}$: 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.
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}}. $