Basic differentiation questions

[computerex]

Homework Statement

Hello guys. I had to do some test corrections for my AP Calculus AB class. I have completed all of them besides the four below. Can anyone tell me where I go wrong?


1. Differentiate y = (1+cosx)/(1-cosx)

dy/dx = [(1-cosx)(sinx)-(1+cosx)(sinx)]/(1-cosx)^2 (quotient rule)
= [(sinx - sinx cosx) - (sinx + sinx cosx)]/(1-cosx)^2 (distribution)
= (-sinx cosx - sinx cosx)/(1-cosx)^2 (simplification)
= (-2sinx cosx)/(1-cosx)^2 (simplification)

Answer choices:

a. -1
b. -2 cscx
c. 2 cscx
d. (-2sinx)/(1-cosx)^2

Choice D is the closest to my answer, however my answer multiplies cosx with the -2sinx.

2. Differentiate y = sin(x+y)
I did this by implicit differentiation:

y' = cos(x+y)[(x)+dy/dx] (doesn't seem correct...)

Choices are:

A. 0
b. [cos(x+y)]/[1-cos(x+y)]
c. cos(x+y)
d. 1

9. Differentiate: y = (secx)^2 + (tanx)^2

y' = [(2secx)(secx tanx)] + [(2tanx)(secx)^2] (product rule)
= 2 ((secx)^2)tanx + 2 tanx (secx)^2
= 2 (sec x)^2 + 3 tanx

Choices are:

a. 0
b. tan x + (secx)^4
c. ((secx)^2)((secx)^2 + (tan x)^2)
d. 4 (secx)^2 tanx

(I skipped Pre-Calculus, which was essentially a trigonometry class, so I had a particularly difficult time with this one)

 

[phyzmatix]

1. Differentiate y = (1+cosx)/(1-cosx)

dy/dx = [(1-cosx)(sinx)-(1+cosx)(sinx)]/(1-cosx)^2 (quotient rule)
= [(sinx - sinx cosx) - (sinx + sinx cosx)]/(1-cosx)^2 (distribution)
= (-sinx cosx - sinx cosx)/(1-cosx)^2 (simplification)
= (-2sinx cosx)/(1-cosx)^2 (simplification)

d/dx cosx = - sinx NOT sinx

2. Differentiate y = sin(x+y)
I did this by implicit differentiation:

y' = cos(x+y)[(x)+dy/dx] (doesn't seem correct...)

Didn't check this one properly, but d/dx x = 1 NOT x

9. Differentiate: y = (secx)^2 + (tanx)^2

y' = [(2secx)(secx tanx)] + [(2tanx)(secx)^2] (product rule)
= 2 ((secx)^2)tanx + 2 tanx (secx)^2
= 2 (sec x)^2 + 3 tanx

Check step 3.

Hope that helps!