Calculus Help Please ASAP?

Question

I'm stuck on these few problems. Somebody please help me I've been struggling for an hour now.

I need to find the derivitive:

1. f(x) = (2x^2 + 5)^7
2. f(x) = 1 / 3 square root of 3 - x^3
3. y = x^3 square root of x + 1

this one need you need to find the tangent line equation
4. f(x) = square root of 2x + 1 (4,3)

this one you need to find the acceleration
5. f(x) = square root of x^3 + 1 with x = 2

all answers/help/suggestions are appreciated
Thank You

Answer 1

Use chain rule for all these problems.
1. f'(x) = 7(2x^2 + 5)^6 (4x)
2. You need to use parenthesis to make it clear.
3. f'(x) = 3x^2 sqsrt(x+1) + x^3/[2sqrt(x+1)]
...

Answer 2

1. Remember that the derivative of a function x^n is n*x^(n-1).
for example, the derivative of x^8 is 8x^7 , the derivative of x^41 is 41x^40.
Second remember the chain rule, that says you can treat all the stuff in parenthesis as a variable, then multiply by the derivative of the stuff in the parenthesis. I've tried to avoid the formal definition involving composition of functions.
for example, suppose f(x) = (3x^3 + 2)^4 . choose the stuff
inside the parenthesis to be u, like, u=3x^3+2
then f(x) = u^4. f'(x) = 4u^3 * derivative of u ("du/dx").
derivative of u is 3x^2. Putting it all together and replacing u with what it is equal to, you get
df/dx = 4(3x^3+2) * 3x^2.
Now you can do something similar for the real problem.

4. The slope of a tangent line is equal to the slope of the curve at the point of tangency. And the slope of the curve is just the derivative of the function at that point. In this case, the point is x=4 y=3. So take the derivative of the function, plug in the value 4, and find the slope of the tangent line. Now just find a line of that slope going through (4, 3).

5. Assuming that f(x) is the formula for the position of x, the velocity will be the first derivative of the function, and the acceleration will be the second derivative of the function.

Sorry to not just give you the answers, I want you to learn how to do it on your own.

If you look like your avatar, you should have no trouble finding a guy in your class very motivated to help you with this stuff.