# Quotient rule

1. Oct 1, 2009

### fghtffyrdmns

1. The problem statement, all variables and given/known data
For some reason, I cannot seem to get the derivative for this.

2. Relevant equations

$$f(a) =\frac {(2500+0.2t)(1+t)}{\sqrt {0.5t+2}}$$

3. The attempt at a solution

$$f(a) =\frac {(2500+0.2t)(1+t)}{\sqrt {0.5t+2}}$$

$$\frac {(2500+0.2t)}{\sqrt {0.5t+2}} (1+t)$$

From here I would just use the quotient rule but I keep getting the wrong answer and I have no idea why.

2. Oct 2, 2009

### RPierre

I'll simplify this for you. See if you can solve the question from here:

First, FOIL the numerator. Next, solve all of our components separately, so:

h'(x) = $$\frac{d}{dx}$$ (0.5t + 2)1/2

= $$\frac{1}{2}$$(0.5t +2)-1/2 * 0.5 <By Chain Rule>

= $$\frac{1}{4}$$ $$\frac{1}{\sqrt{(0.5t+2)}}$$ <By x^-1 = 1/x>

g'(x) = 0.4t + 2500.2 <By power rule x^2 = 2x when the derivitive is taken>

h(x)2 = 0.5t + 2 <x^1/2^2 = x^1>

next, plug in all of that data to the quotient rule, simplify, and see what you get!

3. Oct 2, 2009

### fghtffyrdmns

ahhh, thank you!

at first, I thought I was supposed to treat it separately. Now, I just expanded the numerator and took the derivatives of both sides.

Thank you, sir :).

4. Oct 2, 2009

### Staff: Mentor

Both of you should get your variables straight.
That would be f(t), not f(a).
And that would be $$h'(t) = \frac{d}{dt}(0.5t + 2)^{1/2}$$

Last edited: Oct 2, 2009
5. Oct 2, 2009

### RPierre

I'm not familiar with LaTex and assumed he wasn't using leibniz notation by the prime notation used and the level of calculus, so I simply whipped up a solution with simple principles. I'm sure the original poster got the point as the question was able to be resolved.

Thanks for pointing that out though, I'll be more precise with my answers in the future, I'm also new to these forums.

6. Oct 2, 2009

### fghtffyrdmns

ah yes, it's my fault. I wrote the wrong variables in accidentally.

I got this as my answer

$$\frac {(0.5t+2)^{1/2}(0.4t+2500.2)-(2500+2500.2t+0.2t^{2})}{(2t+8)^{3/2}}$$

Last edited: Oct 2, 2009
7. Oct 3, 2009

### fghtffyrdmns

Hmm. This is not the right derivative. Where did I go wrong?