Question: Given dH(x)/dx = g(x) for all x and that g(x) is a real valued function defined everywhere give an example where g(x) is not differentiable in general.\nAnswer: \ng(x) = 1, for x <= 0\ng(x) = cos x, for x > 0
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i think that is an acceptable answer
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define f(x)=d(x,A)/[d(x,A)+d(x,B)].\nThis is a continuous function f:X->[0,1] (use triangle inequality to show x->d(x,A) is continuous for a closed set A). Also f=0 on A and f=1 on B. Let C=f^(-1)[0,1/3) and D=f^(-1)(2/3,1]. Then C and D are open because inverse images of open sets under continuous functions are open.\n\nNote to Adolfo Boy: It is possible for the distance between two disjoint closed sets to be 0. Consider the x-axis and the y=1/x curve. Your argument does work for compact sets, though. The point (a,b) in your argument need not exist if the infimum is not attained.
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1) If h(x) = f(g(x)) = (x^3 -1)/(x^3 +1), then \nf(x) = (x-1)/(x+1) and g(x) = x^3\n2) If h(x) = f(g(x)) = 3sqrt(x^2 -8), then\nf(x) = 3sqrt(x) and g(x) = x^2 -8
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Problem 1:\nWhich one is a RELATION but not a function\n\nChoices:\n1) {(-10,1),(1,8),(-4,-2)}\nfunction\n2) {-4,1),(1,1),(-2,-2)}\nfunction\n3) {(-5,1),(1,-7),(21,-2)}\nfunction\n4) None\n\n\nProblem 2:\nWhich of the following is a function\n\nChoices:\n1) {(-4,1),(1,8),(-4,-2)}\nrelation\n2) {(-4,1),(1,-8),(-2,-2)}\nfunction\n3) {(-4,1),(1,-8),(1,-2)}\nrelation\n4) None\n`
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What are you trying to do?  You have just defined f(x).  You have not said what to do with it.
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I'm on a roll working with the definition, so I\nthought I'd try this one too. \n\nf(x)=4x/(5x+2)\n\nSubstitute it into the derivative definition:\n\nf'(x)=lim h->0 ((4(x+h)/(5(x+h)+2)) - 4x/(5x+2))/h\n\nMultiply the adjacent terms:\n\nf'(x)=lim h->0 ((4x+4h)/(5x+5h+2) - (4x/(5x+2)))/h\n\nCombine the fractions:\n\nf'(x)=lim h->0 (((4x+4h)(5x+2)-(4x)(5x+5h+2))/((5x+5h+2)(5x+2)))/h\n\nMultiply the adjacent terms:\n\nf'(x)=lim h->0 ((20x^2+20xh+8x+8h-20x^2-20xh-8x)/(25x^2+10x+25xh+10h+10x+4))/h\n\nSimplify:\n\nf'(x)=lim h->0 (8h/25x^2+20x+25xh+10h+4)/h\n\nCancel out the "h" from the top and bottom:\n\nf'(x)=lim h->0 8/25x^2+20x+25xh+10h+4\n\nSince h being zero won't cause an infinity, take the limit:\n\nf'(x) = 8/25x^2+20x+4\n\nBut notice that the denominator can be factored:\n\nf'(x) = 8/(5x+2)^2\n\nYou might notice that this equal to answer #1,\nexcept that it is reduced differently.
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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.
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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.
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1. f(g(x))\n2. g(f(x))
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Definitely not.  Just do the derivative the way you normally would...\n\nRecall, for f(x)=x^n, f'(x) = nx^(n-1), that's how to handle the exponent\n\nuse chain rule...\n\nfor f(g(x)),   f'(g(x)) = f'(g(x))*g'(x), remember...
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If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
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well, to differntiate a funtion using first principle, you have to follow this procedure. \nlet f(x) be a function. then f'(x) is the derivative of f(x) with respect to x. from the first principle, it is given by\nf'(x)=  Lt        f(x+h)-f(x)\n        (h->0)    -------------\n                          h\nwhere Lt represents the limit as h tends to zero.\nnow, take f(x)=cos(x^1/2). so, f'(x) is\n         Lt        cos((x+h)^1/2) - cos(x^1/2)\n        (h->0)    -------------------------------\n                                    h\n    here we apply L'Hopital's rule, because substituting h=0 will yield an indeterminate form. so, the result is \n      Lt        -sin((x+h)^1/2)\n     (h->0)   ------------------\n                   2*((x+h)^1/2)\nsubstitute h=0, we get the result as -1/2*(sin(x^1/2)/(x^1/2))
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There is a defined term oscillation used to describe how much a function fails to be continuous at a point by evaluating the difference between the supremum and infimum of values of f(x) in some non-zero neighborhood of x as the radius of that neighborhood goes to zero. The most this can do is show that f is continuous whenever f(x) = 0, though, and continuity obviously does not imply differentiability (See the absolute value function). I'm guessing that you're therefore just referring to mathematician's interpretation: the tangent line oscillates infinitely near the origin, leaving no derivative.
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Let F(a) be the value of that integral. Take the derivative with respect to a to get\nF'(a)=int x^a * log(x)/log(x) dx\n=int x^a dx\n=1/(a+1) [1^(a+1) -0^(a+1)]\n=1/(a+1).\n\nNow take the anti-derivative to get\nF(a)=log(a+1) +C.\nBut, clearly if a=0, we are integrating 0, so C=0.\nThis gives the result.\n\nThe only point at issue is whether differentiating under the integral sign is legitimate. In spite of appearances, the function inside the original integral is continuous (or can be extended to be continuous) and clearly the partial derivative is continuous in both x and a, so the result follows.
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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
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Given 2 functions f and g with the domains Df and Dg respectively.\n\nAny composite function of f and g will as domain have Dfg=Df ∩ Dg &#92;{ those interval or point where the composite functions has poles coming from the way the functions is composited}\n\nEX 1)\nf=x^2 ; Df={x in R l 0<x<20} by own choice\ng=x ..; Dg={x in Rl x belongs to [-5,5[} by own choice\n\na composite function fg=f/g+1/f-g\nfinding the domain Dfg:\nDfg={ x in R l x belongs to Df ∩ Dg &#92; {those interval or point where the composite functions has poles coming from the way the functions are composited} }\n\nDf∩Dg=\n{x in R l 0<x<20}∩{x in Rl x belongs to [-5,5[}=\n{x in R l 0<x<5}\n--------------------\n\n\nfinding: \nPole={those interval or point where the composite functions has poles coming from the way the functions are composited}\n\nlook at the composite function: fg=f/g+1/f-g.\nWe easily see the composite function has pole for g=0=x => x=0 and for f-g=0=x^2-x=x(x-1)=> x=0 or x=1=>\nPole={ x in R l x=0 or x=1}\n----------------------------------\nnow we can find the domain for the composit function fg\n\nDfg=Df ∩ Dg &#92; Pole ={x in R l 0<x<5} &#92; { x in R l x=0 or x=1} }= {x in R l 0<x<5 & x≠1}\n=======================
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Even 'tho it's obvious, you really should force yourself to say 'derivative with respect to x' because, later on, it won't be quite so obvious what the derivative is being taken with respect to ☺\n\nFirst of all √(4-x²) = (4-x²)^½.  So\n(d/dx)(4-x²)^½ \n½(4-x²)^(-½) (d/dx)(4-x²) (derivative of an expont)\n½(4-x²)^(-½) [(d/dx) 4 - (d/dx)x²] (derivative of difference)\n½(4-x²)^(-½) [0 - 2x] (derivative of constant and exponent)\n½(4-x²)^(-½) (-2x)  (algebra)\n-x(4-x²)^(-½)  (algebra)\n-x/(4-x²)^½ (algebra)\n1x/√(4-x²) (more algebra ☺)  which is the final answer.\n\nSee why you needed to learn algebra? ☺\n\n\nDoug