Recent content by MercuryRising

are you suggesting trig substitutions? i thought you can only do that when the integral is in the form A^2 + V^2

a) find the area bounded by the polar curves r=3sinx, r=1+sinx first i find the points of intersection, x=pi/6, and x=5pi/6 so to find the area i set up the integral A= 1/2 integral from pi/6 to 5pi/6 [(3sinx)^2 - (1+sinx)^2]dx using trig identities, it simplifies to 1/2 the...

the function is 13ln (exp[(x^2+2)] - exp[ln(13x^2+26)]) NOT seperately as in (13ln[exp[(x^2+2)] ) - (exp[ln(13x^2+26)]) sorry if i was unclear, writing in laTex is too tedious for me

hmm how exactly do you get to the algebraic expression 13(x^2+ 2)- (13x^2+ 26) the function dave posted was the farthest i got..:confused:

Im stuck on these probelms Simplify 13ln[exp[(x^2+2)] - exp[ln(13x^2+26)]] exp[ln(13x^2+26) cancels out to 13x^2+26 but i don't see how that helps with the entire probelm Derive x^(e^x) my friend suggested power rule...but that seems a litle too simple and would create a quite a mess thank you...

ahh...i didnt know you can work both sides like that, thank you all very much :biggrin:

If dy/dx = 2y^2 and if y=-1 when x=1 then when x=2, y=? I have no idea how to even start this probelm, i never seen this on any previous practice tests i thought about using the equation y+1 = dy/dx (x-1), but that's probably wrong and not going to help...

well, in this case f(x) = pi so x= pi 1/ f'(pi) = 1/ 1+ cos( pi) = 1/ 1-1 = 1/0 :confused:

ok... so g'(x) = 1/ 1 + cos(g(x)) now they want g'(pi)...and I am stuck again, i can't find any trig identities that can help me in g'(x)

ok.. so f(g(x)) = g(x) + sin(g(x)) = x to find g'(x) in terms of g(x) i did g'(x) + (g'(x))cos(g(x)) = 1..wait that doesn't seem right

since it seems like that finding the inverse is beyond high school calculus..maybe my approach to the probelm was incorrect the probelms states f(x) = x + sin(x), g(x) is the inverse of f(x) so f(g(x)) = x a) write an expression for f(g(x)) in terms of g(x) and b) use f(g(x))= x to find g'(x)...

find the inverse of y= x + sinx i got as far as x=y+siny..but how do i get to y= ____? thanks

hmm, how exactly do i differenciate -(y-2x)+(2y-x)y'=0 ? i get y' and y''s in the derived equation...im so lost..

the probelm told me to take the 1st and 2nd derivatives of x^2-xy+y^2=9 for the 1st derivative i got dy/dx = y-2x/2y-x but I am stuck on taking the second derivative for impicit differenciation =(

sorry, i was in a rush so i posted it in the wrong section :biggrin: 1) i tried (2x/2x) (sin2x)/sin(5x) -> (2x/sin5x) (sin2x)(2x)..now i don't know how to get rid of the (2x/sin5x) 2) hmm...the derivative i got was (X^2-1)/x^2. do i just plug in a random number to get a slope then use...