Calculus, derivatives. Please help! Show work, if possible. Thanks! :)

Answer:

y=-x+pi/3+sqrt(3)/2

Step-by-step explanation:

*I assume you are asking for help on part a because the other question is a calculator question.*

We want to find tangent line to y=2sin(x)cos(x) at x=pi/3.

First I prefer to use y=sin(2x). (Double angle identity: sin(2x)=2sin(x)cos(x) .)

Differentiate using chain rule:

y'=2cos(2x)

y' gives us the slope for any tangent along our curve.

So at x=pi/3, the slope is 2cos(2×pi/3)=2cos(2pi/3)=2(-1/2)=-1.

We also need to know a point our line to find the equation of our line.

At x=pi/3, y=sin(2×pi/3)=sin(2pi/3)=sqrt(3)/2.

So a point on our line is (pi/3, sqrt(3)/2).

So by point-slope form of a line we have

y-y1=m(x-x1)

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

Distribute:

y-sqrt(3)/2=-x+pi/3

Add sqrt(3)/2 on both sides:

y=-x+pi/3+sqrt(3)/2

Answer:

y= -x+(pi/3)+(sqrt3)/2

Step-by-step explanation:

The given is y=2sin(x)*cos(x)

to take the derivative we will need to use the constant rule (which says when taking the derivative of any constant multiplied by an expression you take the derivative of the function and multiply it by the constant) and the multiplication rule to get

y'=2(sin(x)*-sin(x)+cos(x)*cos(x))

this would simplify to be

y'=-2sin^2(x)+2cos^2(x)

We then substitute in the given x value to find the slope of the tangent line at that point and get

-2sin^2(pi/3)+2cos^2(pi/3)

which simplifies to

-2*3/4+2*1/4

which is -1

Since we have the slope we need to find the y value to go with the original x so that we have a point to use in point slope form. When we substitute pi/3 into the given equation we get (sqrt 3)/2 and have the point (pi/3, sqrt3/2)

now we write the point slope form equation

y-(sqrt3)/2= -(x-pi/3)

y-(sqrt 3)/2 = -x+pi/3

y= -x+pi/3+(sqrt3)/2

you could simplify the last two terms, but I think it is simplest to understand this way.

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