Mid Term Help Needed (Limits, Tangent Lines, Differentiation, and Integration)

So I'm going through my Mid Term review and I continue to come across the same problems I've been dealing with on past tests and such.

It's going to be a multiple choice test, but when I come to tangent lines on a MC test I just plug things in and test them with the calculator...That's not a good idea so I'm wondering what exactly is the method for getting the tangent line?

The problem I have is f(x) = x(1-2x)^3 , find the tangent line at point (1, -1) and the answer is y = -7x+6

I don't understand how to get that answer.

~~ And another general question:

When you are taking the derivitive of trigonometric values, are there separate rules for them compared to before? Like, what would the derivitive of y = tan(x) - cot(x) be?

f(x) = x (1-2x)^3

1. find the derivative

f '(x) = (1 -2x)^3 + x (3) (1 -2x)^2 (-2)
f '(x) = (1 -2x)^3 - 6x (1 -2x)^2 [Product Rule]

2. You are looking for the tan. line at (1, -1) so you plug in x =1 into this equation

f ' (-1) = -7

3. Use the formula to find equ ofa line

y - y1 = m (x - x1)

y - (-1) = (-7) (x -1)
y + 1 = -7x + 7

y = -7x +6


y = tan(x) - cot (x)
y' = (sec(x))^2 + (csc(x))^2



Hope this Helps.

differentiated: tanx = sec(x)^2
cotx = -csc(x)^2
sinx = cosx
-cosx = sinx

Can anyone complete this chart? with positive and negative for the values?

Everything in that post helped, but when differentiating, I'm not quite sure what to think of trig identities. Up there is all I know at this point x.X

In gonna be following this format..
f(x) = f ' (x)

Therefore...

sin(x) = cos(x)

cos(x) = -sin(x)

tan(x) = (sec(x))^2

csc(x) = -csc(x) cot(x)

sec(x) = sec(x) tan(x)

cot(x) = -(csc(x))^2

While you explained the notation you were using, so I can't complain too much, the use of "=" there makes me really uncomfortable. Better to use, say, "=>" so that you are saying that the derivative of sin(x) is cos(x) by
"sin(x)=> cos(x)" rather than "sin(x)= cos(x)"!