1. ## Are my book answers wrong?

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=2sinx+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

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?

THANKS!

2. I'll just do the first one real quick

d/dx cos x sin 2x use product rule

d/dx uv = v du/dx + u dv/dx

u = cos x, v = sin 2x
du/dx = -sin x

Use chain rule for d/dx sin 2x
dy/dx = dy/du du/dx
u = 2x
d/dx sin 2x = 2cos 2x

d/dx cos x sin 2x = sin 2x * -sin x + cos x * 2 cos 2x

f'(x) = 2cos 2x cos x - sin 2x sin x

3. Originally Posted by kmjt

2. Find the equation of the line that is tangent to the curve y=2cosx+4cosx at x=pi/4.
$y= 2\cos x +4\cos x= 6\cos x \implies \frac{dy}{dx}= -6\sin x$

The tangent then becomes

$y-f\left(\frac{\pi}{4}\right) = f'\left(\frac{\pi}{4}\right) \left(x - \frac{\pi}{4}\right)$

4. for number three the slope is 2 (not -2)

sin 3pi/2 = -1 not 1

5. Originally Posted by kmjt
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

Where did I go wrong?
d/dx 2 sin x cos x = 2 d/dx sin x cos x

Product rule d/dx uv = v du/dx + u dv/dx

u = cos x, v = sin x
du/dx = -sin x, dv/dx = cos x

so

2 d/dx sin x cos x = sin x * -sin x + cos x * cos x = 2(cos^2 x - sin^2 x)

6. ## Where you have gone wrong

Originally Posted by kmjt
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

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?

THANKS!
1. SIN2X=0 WHEN X=PI/2
2. aRE YOU SURE YOU QUOTED THE QUESTION CORRECTLY?
Y=2COSX+4COSX WOULD SIMPLY BE Y=6COSX
3. SIN(3PI/2)=-1 SO Y=1/2 AND Y'=2
4. yOU MUST USE THE PRODUCT RULE WHEN DIFFERENTIATING HERE GIVING Y'=2COS^2X-2SIN^2X ALTERNATIVELY WE CAN SAY Y=SIN2X SO Y'=2COS2X
5. HERE YOU NEED THE CHAIN (FUNCTION OF A FUNCTION)( RULE GIVING Y'=-6COS^2XSINX
Hope this helps

7. Originally Posted by kmjt
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?
$sin(2x)\ne 2 sin(x)$ and $cos(2x)\ne 2cos(x)$!

With $x= \pi/2$ $2x= \pi$ and $sin(2x)= 0$, not 2. Similarly, $cos(2x)= -1$, not 0.

8. For number one.. my derivative looks like -2sinxsin2x +4cosxcos2x, isn't sin(pi/2)=1? At 90 degrees, y=1, r=1.. and if sin is y/r = 1/1 = 1. How are you getting 0?

For 2.. I made a mistake in the question I'm actually asking
2. Find the equation of the line that is tangent to the curve y=2sinx+4cosx at x=pi/4.

I've completed 3.

I've completed 4.

I think I completed 5.. is this correct what i've shown my work?:

Did I do that right.. when you multiply [(3)(square root of 3)]/4 by pi/3, the 3s in numerator and denominator cancel each other out like that?

9. for 1 the correct derivative is f'(x) = 2cos 2x cos x - sin 2x sin x. See my first comment for steps

2 is a straight foward derivative 2 cos x - 4 sin x

2 cos pi/4 - 4 sin pi/4 = (y - y(pi/4))/(x - pi/4)

2 cos pi/4 - 4 sin pi/4 = (y - 2 sin (pi/4) - 4 cos (pi/4))/(x - pi/4)

y = (2 cos pi/4 - 4 sin pi/4)(x - pi/4)/(2 sin (pi/4) + 4 cos (pi/4))