What is the derivative of sin2x / (sin2x + secx) with respect to x?

[RaptorsFan]

Homework Statement

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

Homework 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

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?

 

[Mark44]

Homework Statement

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

Homework 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

The Attempt at a Solution

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

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

= [(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?

 

[RaptorsFan]

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

 

[Mark44]

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

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

 

[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?

 

[Mark44]

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

Also, in one of your lines you had a numerator of
[(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)]
......^
......|

 

[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?

 

[Mark44]

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?

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

 

[Mark44]

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.

 

[RaptorsFan]

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