# Thread: Need help with finding derivatives

1. ## Need help with finding derivatives

y= sqrt[x] / 2sqrt[x]+3

y=-2sqrt[x^2+4x]

2. Can you use any method to solve or do you need to use the fundamental theorem?

3. Is the first one $\displaystyle y = \frac{\sqrt{x}}{2\sqrt{x}} + 3$ or $\displaystyle y = \frac{\sqrt{x}}{2\sqrt{x} + 3}$?

4. Originally Posted by Prove It
Is the first one $\displaystyle y = \frac{\sqrt{x}}{2\sqrt{x}} + 3$ or $\displaystyle y = \frac{\sqrt{x}}{2\sqrt{x} + 3}$?

5. have you tried the quotient rule for derivatives?

6. Well first of all, since you're dividing by ALL of $\displaystyle 2\sqrt{x} + 3$, you needed to write it as sqrt[x]/(2sqrt[x] + 3)...

Now you should apply the Quotient Rule, namely if $\displaystyle u = u(x)$ and $\displaystyle v = v(x) \neq 0$, then

$\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\,\frac{du}{dx} - u\,\frac{dv}{dx}}{v^2}$.

7. yeah im just having trouble with the square roots though . like do i get rid of them ?

8. Remember that $\displaystyle \sqrt{x} = x^{\frac{1}{2}}$. Then you can apply the power rule for derivatives...

BTW, since this is clearly a Calculus question, why have you posted in Pre-Calculus?

9. thanks for the help

10. taking the derivative of functions with $\sqrt{x}$ in them are much easier if you remember that its derivative is $\dfrac{1}{2\sqrt{x}}$.

quotient rule ...

$y' = \dfrac{(2\sqrt{x}+3) \cdot \dfrac{1}{2\sqrt{x}} - \sqrt{x} \cdot \dfrac{1}{\sqrt{x}}}{(2\sqrt{x}+3)^2}$

multiply stuff in the numerator and combine like terms ...

$y' = \dfrac{\dfrac{3}{2\sqrt{x}}}{(2\sqrt{x}+3)^2}$

clean it up ...

$y' = \dfrac{3}{2\sqrt{x}(2\sqrt{x}+3)^2}$

11. Originally Posted by skeeter
taking the derivative of functions with $\sqrt{x}$ in them are much easier if you remember that its derivative is $\dfrac{1}{2\sqrt{x}}$.

quotient rule ...

$y' = \dfrac{(2\sqrt{x}+3) \cdot \dfrac{1}{2\sqrt{x}} - \sqrt{x} \cdot \dfrac{1}{\sqrt{x}}}{(2\sqrt{x}+3)^2}$

multiply stuff in the numerator and combine like terms ...

$y' = \dfrac{\dfrac{3}{2\sqrt{x}}}{(2\sqrt{x}+3)^2}$

clean it up ...

$y' = \dfrac{3}{2\sqrt{x}(2\sqrt{x}+3)^2}$
Not necessarily... it's just another formula to remember ><