# Homework Help: Calculus I - Derivitive

1. Oct 26, 2009

### RaptorsFan

1. The problem statement, all variables and given/known data

sin2x / (sin2x + secx) d/dx = ?

2. Relevant equations

d/dx (f/g) = (g d/dx f - f d/dx g) / g^2
d/dx sec u = sec u tan u d/dx u
d/dx cos u = -sin u d/dx u
d/dx sin u = cos u d/dx u

3. The attempt at a solution

d/dx y= [(sin2x + secx) d/dx (sin2x) - sin2x d/dx (sin2x + secx)] / (sin2x + secx)^2

= [(sin2x + secx) cosx d/dx (2x) - sin2x d/dx (sin2x) + d/dx secx)] / (sin2x + secx)^2

= [(sin2x + secx) cosx (2) - sin2x cos2x (2) + (sec2x*tan2x) 2] / (sin2x + secx)^2

= [2(sin2x + secx) cosx - 2sin2x cos2x + 2(sec2x*tan2x)] / (sin2x + secx)^2

^ That is my final answer, can anyone confirm for me?

2. Oct 26, 2009

### Staff: Mentor

Mistakes in next line. d/dx(sin2x) = cos2x d/dx(2x) = 2 cos2x.
d/dx(secx) = secx * tanx

Last edited: Oct 26, 2009
3. Oct 26, 2009

### RaptorsFan

On our given formulae sheet sec x is equal to sec x * tan x :S

4. Oct 26, 2009

### Staff: Mentor

OK, my mistake, which I have fixed, but your formula sheet should say d/dx(secx) = secx * tan x, not sec x = secx * tanx.

5. Oct 26, 2009

### RaptorsFan

Yes sir, that is correct. But I am skipping a step.

d/dx (sin2x + secx)

is equal to d/dx sin2x + d/dx secx

I just failed to put that step in, would you agree with this?

6. Oct 26, 2009

### Staff: Mentor

Yes, I agree with that, but what do you get for d/dx sin2x + d/dx secx?

[(sin2x + secx) cosx d/dx (2x) - sin2x d/dx (sin2x) + d/dx secx)]
You're missing a left parenthesis somewhere; it should be right between -sin2x and d/dx. IOW, like so: [(sin2x + secx) cosx d/dx (2x) - sin2x (d/dx (sin2x) + d/dx secx)]
...............................^
...............................|

7. Oct 26, 2009

### RaptorsFan

Thank you.
d/dx sin2x + d/dx secx = cos2x (d/dx 2x) + tanx*secx

Ahh, I see my fault.. the correct solution would be:

= [2(sin2x + secx) cosx - 2sin2x cos2x + (secx*tanx)] / (sin2x + secx)^2

Thank you, I always make stupid mistakes. Is this the answer you are thinking on?

8. Oct 26, 2009

### Staff: Mentor

No, parts are wrong. Here's what I get:
d/dx[sin2x/(sin2x + sec x)] = [(sin2x + secx)(2cos2x) - (sin2x)(2cos2x + secxtanx]/(sin2x + secx)^2
= [2sin2x cos2x + 2secx cos2x - 2sin2x cos2x - sin2x secx tanx]/(sin2x + secx)^2

9. Oct 26, 2009

### Staff: Mentor

A piece of positive criticism about your work is that you are setting things up correctly when you start in on the quotient rule. This makes it very easy to follow what you're doing. A little more practice and you'll be able to do this with much fewer mistakes.

10. Oct 27, 2009

### RaptorsFan

Ahh yes I understand, just working it out on paper for myself now. Thank you