Hmm, the picture is very small.
Hi, this math sheet has been giving me a lot of problems lately. I was wondering if someone could look over it and help me out. I have done most of it, however the questions i have on the attached document i cant get a solid answer. Please give me a hand! THANK YOU VERY MUCH!
Here are hints on how to do the questions:
1) (A) use the product rule:
Product Rule: (ab)' = a'b + ab'
use a = tan^-1(2x) and b = (cos(3x))^(1/2)
you know how to find the derivatives of each of those correct? you will need the chain rule for each
Chain rule: [f(g(x))]' = f'(g(x)) * g'(x)
1) (B) Use the chain rule on each. if you still have problems, say so
1) (C)
let sin^-1(2x) = theta
=> 2x = sin(theta)
draw a right-triangle with one acute angle called theta, and the side oppiste to that angle call 2x and the hypotenuse call 1, use pythagorean's theorem to find the adjacent side. then use the fatc that tan(sin^-1(2x)) = tan(theta) = opposite/adjacent. so you can change t to an algebrac function, then find the derivative normally.
2) find y' for y = xcosx, and plug in x = 2^R in the result. that will give you your m. to find b, plug in 2^R for x in the original equation and then use the coordinates (x,y) = (2^R, y(2^R)) in the point-slope form
3) the area of a trapezium is half the sum of the two parallel sides times the distance between them. write out this formula and use substitutions to get it into one variable. find the derivative and set it equal to zero, this will give you one coordinate for the max value.
6) (A) use the substitution u = 1 - cos(2x)
(B) use substitution u = tanx
(C) use the fact that sin(2x) = 2sinxcosx
so (sin(2x))/cosx = (2sinxcosx)/cosx = 2sinx
post your solutions when you're done so we can examine them. if you still have problems just say so