(Just for someone interested and ignorant.. the best site for detailed solutions would be wolfram-alpha)
First Problem
Differentiate xx
Answer: We do it by chain rule assume it as fxgx so we apply chain rule on this function [d fxgx/d(fx)]x d(fx)/dx + [d fxgx/dgx ]x d(gx)/dx
hence the answer would be (1+log(x)) xx
(Just noticed another way, this one is not brute force but rather smart way)
y = xx
take log
log y = xlogx
now differentiate
dy/y = (logx + x/x)
=> dy = y(logx +1)
same as above ( Maths is consistent when correct :) )
Second problem
Integrate cos2(x) and cos3(x).
cos(2A) = 2cos2(x)-1
=> cos2(x) = [1+cos(2A)]/2
rest is pretty simple
for cos3(x) write it as cos(x)cos2(x) now write squared terms in form of sin and then put sin(x) as y then cos(x)dx would be dy so we will have Integral(1-Y2)dy
rest is simple for this one as well.
Third problem
What is 1.06 raised to the power 10.5
let y = 1.06^10.5
then ln y = 10.5X ln(1.06)
ln(1+x) when x tends to zero is equal to x
hence ln(1.06) can be approximated to 0.06
hence
lny = 0.63
hence y = exp(0.63) which could be approximated around 1.8
(this was the first method that came to my mind)