# Math Help - [SOLVED] derivative

1. ## [SOLVED] derivative

f(x)= 1/sq rt(x) + 2
find y '

is this answer -1 and if so if the sq root had x^2 under it would the answer be -2

2. Which is correct?

1. $f(x) = \frac{1}{\sqrt x} + 2$

2. $f(x) = \frac{1}{\sqrt x + 2}$

3. number 2

4. $f(x) = \frac{1}{\sqrt x + 2}$

$f'(x) = \frac{(\sqrt x + 2)(0) - (1)(\frac{1}{2\sqrt x})}{(\sqrt x + 2)^2}$

$f'(x) = \frac{\frac{-1}{2\sqrt x}}{(\sqrt x + 2)^2}$

$f'(x) = -\frac{1}{2(\sqrt x)(\sqrt x + 2)^2}$

5. how did you get the 1/ 2sqrt(x) in the numerator, do you chand the sq rt to x^2

6. The derivative of a function in the form $\frac{f(x)}{g(x)}$ is $\frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{(g(x))^2}$. This is known as the quotient formula for derivatives.

7. yea i was trying to figure out the process step by step on how sq rt (x) equals 2sq rt(x)

8. The derivative of $\sqrt x + 2$ is given by the power rule:
$\frac{d}{dx} x^n = nx^{n-1}$.

What you have is $x^{1/2} + 2$.

Taking the derivatives:

$\frac{d}{dx} x^{1/2} = (1/2)x^{-1/2} = \frac{1}{2\sqrt x}$

$\frac{d}{dx} 2 = 0$

$\frac{d}{dx} (\sqrt x + 2) = \frac{1}{2\sqrt x}$