I've been doing some questions that I thought I was doing right but then I check the book answer and i'm wrong..:

**1. Find the slope of the function y=2cosxsin2x at x=pi/2.**
So first I differentiated:

y' = (-2sinx)(sin2x) + (2cosx)(cos2x)(2)

= -2sinxsin2x + 4cosxcos2x

Then I determined that sin(pi/2) = 1 and that cos pi/2 = 0 because at 90 degrees, y=1, r=1, x=0.. and sin(pi/2) = y/r = 1/1 = 1.. and cos(pi/2) = x/r = 0/1 = 0 So I subbed these values into my derived equation:

y'(pi/2) = -2(1)(2)(1) + 4(0)(0)(2)

= -4

My answer is the slope is -4, however the book answer says 0. Did I go wrong?

**2. Find the equation of the line that is tangent to the curve y=2cosx+4cosx at x=pi/4.**
So this is what I got when I differentiated y=2cosx+4cosx:

y'=2cosx+4(-sinx)

=2cosx-4sinx

At pi/4: cosx=1/square root of 2, sinx = 1/square root of 2

y'(pi/4) = 2(1/square root of 2) - 4(1/square root of 2)

=0

To get y value:

y(pi/4) = 2(1/square root of 2) + 4(1/square root of 2)

=square root of 2 + square root of 2

y-y1=m(x-x1)

y-(square root of 2+ square root of 2)=0(x-(pi/2))

y=square root of 2 + square root of 2

The teacher said the answer is: y=-(square root of 2)x + (square root of 2 / 4) pi + 3(square root of 2)

Where did I go wrong?

**3. Find the equation of the line that is tangent to the curve y=2cosx-(1/2)sinx at x=3pi/2.**
y'= -2sinx-(1/2)cosx

At 3pi/2: sinx = y/r = 1/1 =1, cosx = x/r = 0/1 = 0

y'(3pi/2) = -2(1) - 1/2(0)

=-2

Finding y value:

y(3pi/2) = 2(0) - (1/2)(1)

= -1/2

y-y1=m(x-x1)

y-(-1/2)=-2(x - 3pi/2)

y=-2x + 3pi + 1/2

Book answer is: 2x + 1/2 -3pi

Why is their 3pi negative, and there 2x is positive?

**4. Find the slope of the line that is tangent to the curve y=2sinxcosx at x=pi/4.**
y'= -2sinxcosx

At pi/4: sinx= 1/square root of 2, cosx= 1/square root of 2

y'(pi/4) = -2(1/square root of 2)(1/square root of 2)

=-1

Book answer: 0

Where did I go wrong?

**5. Find the equation of the line that is tangent to the curve y=2cos^3x at x=pi/3.**
y'= -6sin^2x

At x=pi/3, sinx = (square root of 3) / 2

y'(pi/3) = -6[(square root of 3)/2]^2

I'm not really sure.. how would I work that?

Book answer:

THANKS!