1. The problem statement, all variables and given/known data

1) If f(x)= sin^4x, then f '(pi/3)
2) Given f(x) = x/tanx, find f '(3pi/4)
3) If f(x) = sinxcosx, then f '(pi/6)
4) Differentiate: f(x) = x^2 + 2tanx
Question that I have answered but not sure if it's really the right answer:
5) Find the equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1

2. Relevant equations

Product Rule?
• y= f x g
F'g + g'f

Chain Rule?
• y=f / g
• ((f'g) - (g'f)) / ((g^2))

Slope?
• y=mx+b

3. The attempt at a solution

1)
f(x) = sin^4x then f '(pi/3)
Derivative of sinx = cosx therefore...
I'll assume that sin^4x has the derivative of cos^4x
Now, plug in the number...
cos^4(pi/3) = 0.065

Is that right?

2)
Given f(x) = x/tanx, find f '(3pi/4)
Chain Rule:
((f'g) - (g'f)) / (g^2)
Therefore...
((1 * tanX) - (??? * X)) / ((tanx^2))

Let * be multiplication sign and the ??? to be the "I don't know".
So, I got stuck of what the derivative of "tanx". However, what I do know is that:
tanx = sinx / cosx
tanx' = cox / -sinx <-------Is that right?

If yes, then how am I suppose to make my equation by using "cox / -sinx "?
By plugging that in... I get this:

(((1 * tanX) - ((cox / -sinx) * X))) / ((tanx^2))
Then I'm really stuck on that one... I mean, if I do plug in the "pi/3" to the "x" variables then it will just be a mess. Unless that's the only way to get the answer? Or should have I used the product rule instead?

3)

If f(x) = sinxcosx, then f '(pi/6)
Product rule:
y= f x g
F'g + g'f
Therefore...
= (cosx*cosx) + (-sinx*sinx)
= (cos^2x) + (-sin^2x)
= (cos^2(pi/6)) + (-sin^2(pi/6))
= 0.633

-Let * be a multiplication sign
-Is that right?

4)
Differentiate: f(x) = x^2 + 2tanx

So, I'll just get the derivative of the equation...
2x + 2(???)

-Let ??? be "I don't know".
So, I'm stuck. I have no idea what's the derivative of tanx have. I already encountered this problem in question #3 and I assumed that it would be:

tanx = sinx / cosx
tanx' = cox / -sinx

Is that right? If yes, then I would get this equation:
2x + 2(cosx/-sinx)

Is that right? If yes, can I simplify it much more?

5)
Find the equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1
y=mx+b
Chain Rule:
= 1(x+1) - 1(x-1) / (x+1)^2
= x+1 -x +1 / (x+1)^2
= 2 / (x+1)^2
= Plug in "x"
= 2 / (1+1)^2
= 2 / 4
= 1/2
slope (m) = 1/2

Now that I have the slope, I'll just get the x & y values from plugging in "1" to the equation.
y = (1-1) / (1+1)
y = 0/2
y = 0
So: x = 1 and y = 0
Then reflect on the slope equation:
y = mx+b
Plug in the numbers from what I have gotten before:
0 = 1/2(1)+b
0 - 1/2 = b
-1/2 = b

So...
y = 1/2(x) + (-1/2)
I'll multiply the whole equation by "2" to make it more neater.
2y = 2(1/2x) + 2(-1/2)
2y = x -1

Is that right?

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I know my solutions were kind of long. But I hope that you could help me. I really want to make this happen. Or at least answer the questions correctly or in a much simplified way. =)