# Basic differentiation questions

by computerex
Tags: basic, differentiation
 P: 68 1. The problem statement, all variables and given/known data 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)
P: 364
 Quote by computerex 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!

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