f(x) = g(x)(h(b) - h(a)) - h(x)(g(b) - g(a)) where g and h are differentialbe on (a,b)  use that to show:\ng'(y)(h(b) - h(a)) = h'(y)(g(b) - g(a)) for some point y in (a,b)
135439
Use Rolle's theorem? :-)\n\nRolle's Theorem\nhttp://mathworld.wolfram.com/RollesTheorem.html\nLet f be differentiable on the open interval (a,b) and continuous on the closed interval [a,b]. Then if f(a) = f(b), then there is at least one point c in (a,b) where f'(c) = 0.\n\nGiven: f(x) = g(x)(h(b) - h(a)) - h(x)(g(b) - g(a))\nwhere g and h are differentiable on (a,b)\n{and continuous on [a,b] -- differentiability on (a,b) only implies continuity on (a,b)}\n\nThen f(a) = g(a)(h(b) - h(a)) - (h(a)(g(b) - g(a))\n = g(a) h(b) - g(a) h(a) - g(b) h(a) + g(a) h(a)\n = g(a) h(b) - g(b) h(a)\n\nand f(b) = g(b)(h(b) - h(a)) - h(b)(g(b) - g(a))\n = g(b) h(b) - g(b) h(a) - g(b) h(b) + g(a) h(b)\n = g(a) h(b) - g(b) h(a)\n = f(a)\n\nSo f(a) = f(b) therefore there is a y in (a,b) -- called c in the statement of Rolle's theorem above -- such that f'(y) = 0 = g'(y)(h(b) - h(a)) - h'(y)(g(b) - g(a)) and so at that point y you have g'(y)(h(b) - h(a)) = h'(y)(g(b) - g(a)) as desired.\n\nSee http://cr.yp.to/papers/calculus.pdf
621143
f'(x) = d(22)/dx + (25) (d (ln x) / dx) + (ln x ) (d(25)/dx)\nf'(x) = 0 + 25 (1/x) + (ln x)(0)\nf'(x) = 25/x
472349
I'll call g(x) = (integral from 0 to x) f(s) ds.\nWe just need to see what g(-x) is. \ng(-x) = (integral from 0 to -x) f(s) ds.\nIf f is even, then f(s) = f(-s), so this equals:\n(integral from 0 to -x) f(-s) ds\nand that is equivalent to\n(integral from 0 to x) f(s) ds = g(x), so g is even.\nSimilarly, if f is odd, f(s) = -f(-s), so it equals\n(integral from 0 to -x) -f(-s) ds\nwhich is\n(integral from 0 to x) -f(s) ds = -g(x), so g is odd.
551054
if Q is a function of x then it should be something like this:\n\nfactor it all out:\nkxQ-kx^2\n\nso when you derive Q, you have to treat it as a function.\n\nkQ + kxQ' - 2kx\n\nsolve for Q'\n\ndQ/dx = Q' = (2kx-kQ)/kx\n\ni hope this is the correct method you are talking about.
530419
f'(x)=(4/3)x^(1/3) +(4/3)x^(-2/3)
198209
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
194207
If it is a linear differential equation like\n\na(x)y'(x)+b(x)y(x)=f(x)+g(x)\nthen solve separately\naz'+bz=f\nand\naw'+bw=g\n\nThe sum z(x)+w(x) of such two "partial solutions" will give you a particular solution of your equation, as\na(x)(z+w)'(x)+b(x)(z+w)(x)\n=f(x)+g(x)\n\nP.S.\nIt seems you have not a sum of functions at the right side, but a product: (e^-t)*cost\nIf it is so, you have to see it as a unique function, and search for a particular solution of this form:\n\nx(t)=A(e^-t)cost+B(e^-t)sint\nFind the good constant A and B by taking the derivatives and placing them in the original equation.
406325
The derivative is [(x-4)- (x+1)]/(x-4)^2 = -5/(x-4)^2\n\nIf we use the definition of the derivative:\n\n\ndy/dx = Lim h -> 0 {f(x+h) -f(x)}/h\n= lim h-> 0 [(x+h+1)/(x+h-4) - (x+1)/(x-4)]/h\n                   (x+h+1)(x-4)-(x+1)(x+h-4)\n= lim x-> 0 ---------------------------------------\n                   (x+h-4)(x-4)h\n\n= lim h-> 0 -5h/(x+h-4)(x-4)h = -5/(x-4)^2
451172
It's a product:\n\nif f(x) = a*b\nthen f'(x) = a*(b') + b*(a')\n\nAlso, you must multiply by the derivative of "what's inside."  For this problem, you must take the derivative of "what's inside" the cosine parentheses (2 + 3x - 5x^7).\n\nSo for this one, a = x^5 - 3x^3 and b = cos(2 + 3x - 5x^7).  "What's inside" is (2 + 3x - 5x^7).  Therefore:\n\n--- --- --- --- --- --- --- --- --- ---- ---- --- "what's inside"\nf'(x) = ( (x^5 - 3x^3)(-sin(2 + 3x - 5x^7)*(3 - 35x^6)) + (cos(2 + 3x - 5x^7)*(5x^4 - 9x^2))
393069
1. f(g(x))\n2. g(f(x))
513250
say g(x)=2x^2+3x+8\nh(x)=x-8\ng(x)+h(x)=2x^2+4x=f(x)\n\n2.let f(x)=x^2\nf(-x)=(-x)^2=x^2 we say f(x)=x^2 is an even function\n\n3.let f(x)=x^3\nf(-x)=(-x)^3=-x^3\nf(-x)=-f(x)\nwe say f(x)=x^3 is an odd function
406201
Actually, kaksi_guy seems to be onto something. Unfortunately, not only does his proof have the flaw that you noticed, but even if that were rectified, he would still have only proved that f is zero at infinitely many points on the interval, not that f is zero on every point in the interval. Still, his post is the inspiration for what follows:\n\nFirst, we need to assume that g(x) is defined everywhere on the interval between the two points. We don't need to assume that g(x) is continuous, but if we allow a function that is undefined at even one point, such as g(x)=(x-3/x)/2 (which is undefined at zero), we may easily generate exceptions to the theorem (in this case, f(x)=x²-1).\n\nWith that out of the way, we can begin our proof. Suppose f(x) = 0 at two points a and b. By the extreme value theorem, f(x) has both a maximum and a minimum on the closed interval [a, b]. Let us denote the minimum value f(x) attains on [a, b] as min(f(x)) (and analogously for max(f(x))). Now, since f(a)=f(b)=0, min(f(x))≤0 and max(f(x))≥0. Suppose that min(f(x))<0. Then the minimum does not occur on the endpoints, so it must occur on a point c in the open interval (a, b). It is readily shown that this implies f'(c)=0. Therefore, from the original differential equation, f''(c) - f(c) = 0 and thus f''(c) = f(c). However, f(c)<0, so this implies the function is concave down at c, contradicting the fact that c is a minimum. Thus min(f(x))=0. Similarly, if max(f(x))>0, then there would be a point c on (a, b) which is a maximum, but is concave up, which is also a contradiction. Thus, min(f(x))=max(f(x))=0, and a function whose minimum and maximum are both the same is a constant function, in this case f(x)=0 on the interval [a, b]. Q.E.D.
432505
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
439934
"What i know: it can also be written as (x^2 + cos x) ^ -1/2\nand the derivitive of that = (2x - sin x) ^ -1/2"\n\nWhat you said here was to take the derivative of the inside function, but what you do is take the derivative of the outside function and multiply it by the derivative of the inside function.\n\nIf you have a composition of two functions such as h(x) = f(g(x)), then h'(x) = f'(g(x)) * g'(x) [or as in your notes, f(g) * g']\n\nThat is why you must use the chain rule for 1 and 3 but not for 2, since 2 does not involve a composition:\n\n1. f(x) = (x^2 + cos x) ^ (-1/2)\n\nf(x) = g(h(x)) where g(x) = x^ -1/2 and h(x) = x^2 + cos x\n\n3.  f(x) = 2 ^ (x^2 - sin x)\n\nThis isn't a product- it's an exponential function of another function.\n\nf(x) = g(h(x)) where g(x) = 2 ^ x and h(x) = x^2 - sin x
548515
Here is an outline:\n\nf is an ONTO function. So for EVERY element b in B there is an\nelement a in A so that f(a) = b.  That is what onto means.  All the\nelements in B have some element in A that f maps it into.\n\nSince g is not one to one there must be two elements  in B that get mapped into the same element in C.  Call them b1 and b2. Show that b1 and b2 are part of the codomain of f.  f(x) maps into all B, so there must be a a1 and a2 that get mapped into b1 and b2.  So then we have g(f(a1)) and g(f(a2) that map into the same element, and hence g(f(x)) is not one to one.
529249
g(x)=f(2(x+4)\n=in f(x) whereverx is there sustitute 2(x+4)\nso the answer is 2x+8\n\n2.g(x)=f(-(x-2)\n=in f(x)=x^2 wherever there is x substitute\n-(x-2)\nso [-(x-2)]^2 will be your answer\nwhich is equal to\nx^2-4x+4\n\n3.g(x)=2f(-x)-5\nin f(x)=1/x,substituting\nf(-x)=-1/x\n2f(-x)=-2/x\nso -2/x-5 is your answer
497666
Take the derivative of the original function. For your given example, if I remember my calculus correctly:\nf(x)=y\nf'(x)=1\nGraph the new equation.