1. ## Quotient Rule

Consider g(x)= 3/square root of 7x-5

Determine g' (x) in two ways

1) by using quotient rule
2) by writing as 3(7x-5) to some power (what power?)

2. For #1, simply apply the Quotient Rule. We need to see that you've at least tried it.

For #2, they want you to write $\frac{1}{\sqrt{7x-5}}$ as $(7x-5)^{-1/2}$ and use the Product Rule.

3. ## Thanks but

I know that i have to use the quotient rule and i know that the quotient rule is v(x) * u'(x)-u(x)*v'(x) / [v(x)]^2

However i dont know how to plug the numbers into it and solve it...not sure how to do that with a square root...

4. Do you know that $\sqrt{x} \Leftrightarrow x^{\frac{1}{2}}$?

5. ## I appreciate anyone that can write answer in full

Not trying to take advantage of everyone's generosity..I am new to this and just having an awful time grasping the concept of derivatives and chain rules. Once i have the answer to it, I can review it and it helps me to understand how to move forward with other questions...

6. ## Anyone

7. Originally Posted by dimanre
Don't know how else to explain using the Quotient Rule, other than to just use it. Do you know how to take derivitives? That's all you need to know. If you don't know how, then you have somewhat more significant problems. If your problem is with the algebra, then post your work and we'll show you where you may have gone wrong. But we're not going to do your homework for you.

8. Fair enough -- sometimes seeing a worked example can help with understanding.

$g(x)=\frac{3}{(7x-5)^{1/2}}$

$g'(x)= \frac{0 \cdot (7x-5)^{1/2} - 3 \cdot \frac{1}{2}(7x-5)^{-1/2} \cdot 7}{(7x-5)^1}=$

$\frac{-\frac{21}{2}(7x-5)^{-1/2}}{7x-5}=-\frac{21}{2}(7x-5)^{-3/2}=-\frac{21}{2(7x-5)^{3/2}}$

This clear things up?

9. ## Thank you Scorpion

I really do not expect anyone to do my HW..I simply needed to look at the problem done so that I can use that as my example for other problems. I am just starting out and admittedly over my head..but certainly do not want people doing my HW...thank you very much for helping me..I REALLY appreciate it...