get your 10 point with this calculus question!?
912068
suppose f(x)=x has no real root.\n\nDon't you mean f(x) = 0 has no real root?\n\n\n1) If f(x) = 0 has no real root then ot means that there is no R (R real) such that f(R) = 0 ie f(R) = c where c is real for all R\n\nf(f(R)) = f(c) ≠0 as f(Any Real number) ≠0 from the above statement\n\n2) Suppose that f(f(x)) has a real solution.\nf(f(s)) = 0 where s is real\nNow f(s) = k, a real number\nSo f(k) = 0 which contradicts the original statement that f(x) has no real solutions
893896
y=x^2\ny'=2x\n2x=tan60*\n=rt3\n=rt3/2\nsub y=(rt3/2)^2\n=3/4\nthe point is (rt3/2,3/4)
779791
(f(2) - f(0))/(2 - 0) = 2, but f'(x) < 0 for every x, except x=1 where it is undefined.\n\nThe function has to be continuous in [0,2] and differentiable in (0,2).  For this f(x), it is undefined at x=1, so the conditions of the MVT are not met.  Math is still safe!\n\nCheers!
70409
Go back to math class and figure it out!
847176
f(x,y) = x^2*y*(e^(-x^2-y^2))\n\nfx(x,y) = 2x*y*e^(-x^2 - y^2)(1-2x^2) this is wrong:\nf_x(x,y) = x^2y ( e^(-x^2-y^2)) (-2x) + 2xye^(-x^2-y^2))\n= 2xye^(-x^2-y^2) [-x^2 + 1] =0\nif x=0 or y=0 or x^2 =1, \nx=0, y=0, x=1 or x=-1\n\nfy(x,y) = x^2*e^(-x^2 - y^2)(1-2y^2) \nf_y(x,y)=  x^2 y (e^(-x^2-y^2)) (-2y) +  x^2(e^(-x^2-y^2))\n=x^2e^(-x^2-y^2) [-2y^2 + 1] \nx=0, y=1/sqrt(2),y=-1/sqrt(2)\n\nso the critical points are: (0,y)\n(0,0), \n(1,1/sqrt(2)), (1,-1/sqrt(2))\n(-1,1/sqrt(2)),(-1,-1/sqrt(2))\ns
43939
is the above question in first degreee derivative??
618077
go to www.curnaturecalculus.com  They have all the answers.\n\nSplooch!
263164
I won 10 points!\n\nThis is pretty easy. Try to stump me with harder ones.
1066492
x = -4\n10points thank u!
686844
give me my ten points!!!!!
492355
Check out Stewart's Calculus(refer to the first link). If you are looking for something  more rigorous refer to Apostol's Calculus(refer to second link).
30751
The Edifier has it right, but darned if I can follow the notation. You really need to see this written out like a textbook to understand what's going on.\n\nClick the link below in "sources" to see the answer hosted at my website www.calculus-help.com. Right now, Yahoo! answers, as far as I can tell, doesn't allow graphics, which makes it tricky to answer ANY math questions.\n\nPlease don't post math answers, guys, if the poster knows the answer but can't get there, and you get something completely different.
924339
a) y= 10^(tan (πθ))\n= e^(ln10*tan(πθ))\n= e^u where u = ln10*tan(πθ\ndy/dθ =dy/du * du/dθ\n= e^u * ln10*πsec²(πθ)\n= ln10*πsec²(πθ)*e^(ln10*tan(πθ))\n= ln10.πsec²(πθ)*10^(tan (πθ))\n\nb) y= √[tln(t^4)]\n= √[4t*ln(t)]\n= 2√[t*ln(t)]\n= 2u^½ where u = t*ln(t)\ndy/dt = dy/du * du/dt\n= u^-½ * (ln(t) + 1)\n= (ln(t) + 1)/√[tln(t)]\n\nc) y = arctan(arcsin (x^½))\n= arctan u where u = arcsin v where v = x^½\ndy/dx = dy /du * du/dv * dv/dx\n= 1/(1+ u²) * 1/√[1 - v²] * ½x^(-½)\n= 1/[2√(1 - x)*(1 + (arcsin (x^½))²)]\n\nd)xe^y = y - 1\ne^y + xe^y * dy/dx = dy/dx\nSo dy/dy - xe^y * dy/dx = e^y\nie dy/dx (1 - xe^y) = e^y\ndy/dx = e^y/ (1 - xe^y)\n= 1/(e^(-y) - x)
1108074
10 points just from you!
971776
y=10^x = e^(x ln 10)\nApplying the chain rule:\ndy/dx = e^(x ln 10) * ln 10 = 10^x ln 10\nAnd in general:\nd(a^x)/dx = a^x ln a
1132233
this question will require calcs using logarithm.\nhence will take time
118732
good luck man... http://mathforum.org/library/drmath/sets/high_calculus.html