How to solve the equation with square roots

1. May 2, 2009

a1dogtraining

1. The problem statement, all variables and given/known data
I am having trouble to find out what rule to use when solving this equation:
y=5x^2 square root of (2x^3)/15x^7 square root x

How do you right square roots on the keyboard. As you can see I am quit new to this forum

2. Relevant equations
I have tried to use the chain rule and the quotiant rule. But when I use a different rule I get a different answer.

3. The attempt at a solution

2. May 3, 2009

qntty

Re: Differentiation

Is this the correct equation? (you can see the markup by quoting me)

$$y=5x^2 \sqrt{\frac{2x^3}{15x^7}\sqrt{x}}$$

3. May 3, 2009

a1dogtraining

Re: Differentiation

No
I think it is $$y=5x^2 {sqrt{2x^3}}\15x^7{sqrt{x}}$$

4. May 3, 2009

a1dogtraining

Re: Differentiation

This is what I mean sorry for that

$$y=5x^2\sqrt{2x^3}{frac{15x^7}\sqrt{x}} tex][/QUOTE] 5. May 3, 2009 qntty Re: Differentiation You can differentiate x to a fractional powers the same way you can differentiate x to an integral power. Try simplifying it first. 6. May 3, 2009 a1dogtraining Re: Differentiation Thanks anyway I cann't get the question right. The 5x^2 and sqrt(2x^3) are on the top and 15x^7 and sqrt(x) are on the bottom. lets try that. 7. May 3, 2009 HallsofIvy Re: Differentiation So [tex]\frac{5x^2\sqrt{2x^3}}{15x^7\sqrt{x}}$$?
As qntty suggested, you can write the square root as a fractional power: $\sqrt{x}= x^{1/2}$.

$$\frac{5\sqrt{2}x^2x^{3/2}}{15x^7x^{1/2}}= \frac{5\sqrt{2}x^{2+3/2}}{15x^{7+ 1/2}}= \frac{5\sqrt{2}x^{7/2}}{15x^{15/2}}$$
$$= \frac{\sqrt{2}}{3}x^{7/2}x^{-15/2}= \frac{\sqrt{2}}{3}x^{-4}$$
That should be easy to differentiate.

Last edited by a moderator: May 4, 2009
8. May 3, 2009

a1dogtraining

Re: Differentiation

Thank you very much. Can you just tell me what rule you used, so I can apply it to other equations.

9. May 4, 2009

HallsofIvy

Re: Differentiation

I used a number of "rules". I used the fact that $\sqrt{x}= x^{1/2}$, I used the laws of exponents to reduce the problem to a single power of x, and I used the power rule to differentiate.