Question

1. For f(x)=sin^(2)x and g(x)=0.5x^2 on the interval
[-pi/2,pi/2], the instantaneous rate of change of f is greater than the instantaneous rate of g for which value of x?
a. 0
b. 1.2
c. 0.9
d. 0.8
e. 1.5

2. if tan(x+y)=x then dy/dx=??
a. tan^2(x+y)
b. sec^2(x+y)
c. ln|sec(x+y)|
d. sin^2(x+y)-1
e. cos^2(x+y)-1

Please help !!!

Answer 1

f(x) = sin^2 x or (sinx)^2
f ' (x) = 2sinxcosx or sin (2x)
g ' (x) = x
so is sin (2x) > x for
x = 0, no
x = 1.2 , sin(2.4) = .67.. , so no
x = .9 , sin (1.8) = .97.. which is > .9 , so YES
x = .8 , sin(1.6) = .99957.. which is > .8 , so YES
x = 1.3 , sin 3 = .14 , so no

The graph of both y = sin 2x and y = x
intersect at x = 0 and x = .947747
http://www.wolframalpha.com/input/?i=sin%282x%29+%3D+x
so sin(2x) > x for all values between 0 and .9477..
the two given values of x = .9 and x = .8 fall within that as my calculations show

Answer 2

#2

tan(x+y) = x
then
sec^2 (x+y) * (1+dy/dx) = 1
sec^2 (x+y) + dy/dx sec^2 (x+y) = 1
dy/dx = (1 - sec^2 (x+y) )/(sec^2 (x+y) )
or
= 1/sec^2 (x+y) - sec^2 (x+y)/sec^2 (x+y)
= cos^2 (x+y) - 1

looks like e) is it.