Let  f(x) =e^x  - e^-x  & g(x)= e^x  + e^-x  .Also, let h(x)= f(x)/g(x).  Then... cont.?
816466
Aegor is correct, but here's another way to see it.  Simplify h(x) by multiplying top and bottom of the fraction by e^x; then \nh(x) = [e^(2x) - 1]/[e^(2x) + 1].  (Note careful use of brackets.)  Clearly, as x tends to minus infinity, e^(2x) tends to zero, so h(x) tends to just -1/1, i.e. -1.\n\nLive long and prosper.
749046
well the term you've quoted is a constant...\n\nso (e^2/1+e^2)x would be the answer to that question.\n\nbut i'm guessing you typed it wrong, or are confused.
873165
f(g(x)) = (g(x)+4)^2 = (8x-8+4)^2 = (8x-4)^2 = 64x^2-64x+16\n\nThat's your composite function.\n\nHope this helps.
940656
dy/dx = -(e^x + e^(-x)) / -(y (e^x - e^(-x) + 1)³)\n\nSo ∫ydy = ∫(e^x + e^(-x)) dx/(e^x - e^(-x) + 1)³ \n\nLet u = e^x - e^(-x) + 1 thus du = (e^x + e^(-x) ) dx\n\nSo ∫ydy = ∫du/u³\n\nSo ½y² = -½*1/u² + ½A\n\nie y² = A - 1/u²\n= A - 1/(e^x - e^(-x) + 1)²
935326
You have got to use expansion series!\n\nX->0+ [e-(1+x)^(1/x)]/x =e/2 \nlet x=0+h\n=>as x->0, h->o\n\nh->0 [e-(1+h)^(1/h)]/h\n\nlet y=(1+h)^(1/h)\n=>log y=(1/h)log(1+h)\n=>y=e^[(1/h)log(1+h)]\n\nexpansion of log(1+x)=x - (x^2)/2..................\n=>e^[(1/h)log(1+h)]= e^[(1/h)[h - (h^2)/2!............]\n\n=>e^[(1/h)[-h - (h^2)/2!............]=e^[1-h/2!........]\n\ntherefore,\n\nlet y =lim h->0 [e-(1+h)^(1/h)]/h=lim h->0 {[e- e^(1-h/2!...)]/h}\n\nexpansion of e^x=1+x+(x^2)/2!.................\n\n=>lim h->0 {[e- e^(1-h/2...)]/h}=lim h->0 {-e[e^(-h/2.....) -1]}/h\n\nexpansion of e^x=1+x+(x^2)/2!.................\n=>lim h->0{-e[1+(-h/2-(h^2)/3.....)+ ........-1]/h}\n\n=>lim h->0{-e[(-h/2-(h^2)/2...)+ [(-h/2-(h^2)/2...)^2]/2!+.....]}/h\n\n=>lim h->0{-e(-1/2 -..(terms containing highter powers of h))}\n\n=>e/2\n\n\nP.S.  By the way L' Hospitals rule would fail here as limit is not of form 0/0 or ~/~. It is of form a/0 where a is constant.
953689
A clue:\n\nsin h x = (e^x + e^-x) / 2\n\nCalculate e^(2-i) = e^2 e^-i, etc\n\nsin h x = (e^2 e^-i +e^-2 e^î)^2\n\nAna
532548
I. is not true because if we take x=y=z=1, we get\n   f(1,1,1)=1 > 2 and 2/3=g(1,1,1) which is a false  \n   statement.  \nII. IMHO it is also false because f and h are the same  \n    functions f: R^3 -> R, h(x,y,z)=f(x,y,z)=max(x,y,z)\nIII. |i(x,y,z)|^2=j(x,y,z) would be true (because of the  \n     absolute value) but the statement in III. is again \n     false because if you take eg. point (-1,0,0) you get\n     (i(x,y,z))^2=(min(-1,0,0))^2=-1^2=1 but\n     j(x,y,z)=min(x^2,y^2,z^2)=min(1,0,0)=0 which doesn't  \n     equal 1.\nIV.  if g(x,y,z)=0 then x+y+z/3=0. Of course, you can take \n     such two negative values for two numbers among x,y,z \n     that the equation holds. There can always be a \n     function of maximum t(x,y,z)=max(x,y,z) irrespective \n     of the fact that g=0 !\nQs:  to show that a equation doesn't hold you need to pick\n     such a point (in this case in R^3 so (x,y,z)) that it \n     shows the contradiction. You just need to practice :)  \n     I had such ex. on my first term of studies (maths) in \n     Algebra I. And if a statement is always correct, you \n     have to proof it :)
2024
Case 1: f(x) := x-5*y; (y is parameter)\n\nL{f(x)} = int(exp(-s*x)*f(x),x=0..infinity);\nL{f(x)} = int(exp(-s*x)*(x-5*y),x=0..infinity);\nL{f(x)} = 1/s^2-5*y/s;\n\nCase 2: f(y) := x-5*y; (x is parameter)\n\nL{f(y)} = int(exp(-s*y)*f(y),y=0..infinity);\nL{f(y)} = int(exp(-s*y)*(x-5*y),y=0..infinity);\nL{f(y)} = x/s-5/s^2;
1161844
use this formula:\n\n[ g(h)  -  g(0)  ]    /   [ h  -  0 ]\n\ni.e.\n\n [  2/(h+1)   -   2/(0+1)  ]  / [ h  -  0  ]  =  [ 2/(h+1)   -  2]   /h\n\n=  [( 2 -2h  -2) / (h+1) ]  / h   =  -2/(h+1)\n\nTry it for f(t)
492321
The inverse function of f(x) = e^x is ln(x) because \nln(e^x) = x.  \nWhen x = 1, ln(e^1)=1 ; When x = e, ln(e^e) = e\nThe equation in exponential form is:\ne^ln(e^x) = e^x\n\nIf you have a function g(x) which is the inverse function of f(x), then g(f(x)) = x. For example, arcsine(x) is the inverse function sine(x) and arcsine(sine(x))= x
779760
if by this you mean\n\nf(x) = (x^(2/5)) * 5(x - 1), then\n\nf'(x) = (2/5)x^((2/5) - 1)*(5(x - 1)) + (x^(2/5))(5(x - 1)')\nf'(x) = (10(x - 1)/5)x^((2/5) - (5/5)) + 5x^(2/5)\nf'(x) = (2(x - 1))x^(-3/5) + 5x^(2/5)\nf'(x) = ((2x - 2)/(x^(3/5)) + 5x^(2/5)\nf'(x) = ((2x - 2) + 5(x^(2/5) * x^(3/5)))/(x^(3/5))\nf'(x) = (2x - 2 + 5x^((2/5) + (3/5)))/(x^(3/5))\nf'(x) = (2x - 2 + 5x^(5/5))/(x^(3/5))\nf'(x) = (2x - 2 + 5x)/(x^(3/5))\nf'(x) = (7x - 2)/(x^(3/5))\n\n------------------------------------\n\nBut if you mean\n\nf(x) = (x^2)/(5(x - 1))\n\nf'(x) = (x^2)(5(x - 1))^(-1)\nf'(x) = (2x^(2 - 1))((1/5)(x - 1)^(-1)) + (x^2)((-1/5)(x - 1)^(-1 - 1))\nf'(x) = ((2x)/(5(x - 1)) + ((x^2)/(5(x - 1)^2))\n\nANS : f'(x) = ((2x)/(5(x - 1)) - ((x^2)/(5(x - 1)^2)\n\nor you can do it like this\n\nf'(x) = (2x(x - 1) - x^2)/(5(x - 1)^2)\nf'(x) = (2x^2 - 2x - x^2)/(5(x - 1)^2)\nf'(x) = (x^2 - 2x)/(5(x - 1)^2)\n\nor\n\nf'(x) = (x^2 - 2x)/(5(x - 1)(x - 1))\nf'(x) = (x^2 - 2x)/(5(x^2 - 2x + 1))\nf'(x) = (x^2 - 2x)/(5x^2 - 10x + 5)\n\nANS : (x^2 - 2x)/(5x^2 - 10x + 5)
180190
let f(x) = (x^2) sin(x+2)\nthe derivitive of f(x) [ f'(x) = 2x sin(x+2) + (x^2) cos(x+2)]\n\nThis is done by taking the product rule {g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)( the derivitive of the first times the second plus the derivitive of the second times the first) \nwhere g(x) = x^2 and h(x) = sin(x+2)
838016
This is basically just asking what is f(x)/g(x)\n\nso your answer would be \n\n(x+2) / (3x^2 - 5)
977007
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...
1130821
so for the first part, f(x)=x \nthe derivative no matter what is 1 \n\nfor the second part f(x)=1\nthe derivative no matter what is 0 \n\nso you will have combinational solution
316594
The anti-derivative of x^(2)*e^(x)^(3) is (1/3)*e^(x)^(3)\n\nYou can always check by taking the derivative of your answer.
899001
You didn't write your problem clearly.  I don't know if g(x) is:\n\n1)  x² + (7/x³) - 2, or\n2)  (x² + 7) / (x³ - 2)\n\nThe quality of the answers you receive are in direct relation to the quality of the question you write!  :-)\n\nAssuming you meant (1):\n\ng '(x) = 2x - 21x^-4\n\nAssuming you meant (2):\n\ng '(x) = (-x^4 - 21x² - 4x) / (x³ - 2)²\n\nEDITED TO ADD:\n\nHi Jeff, I got your email but couldn't respond; yahoo said your email was "unconfirmed" whatever that means!  So, I am responding to you here. Hope you see this!\n\nYOU WROTE: "I was wondering if you could explain how to work through this problem"\n\nMY RESPONSE:\n\nHi Jeff, sure I am happy to explain. Basically, when you differentiate a fraction f(x) over g(x) the rule is "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared" -- i.e., \n\n[ (g(x)*f '(x)) - (f(x)*g '(x)) ] / (g(x))^2\n\nSo in your problem f(x) = x^2 - 7\ng(x) = x^3 - 2\n\nSo derivative = [ (x^3 - 2)*(2x) - (x^2 - 7)*3x^2 ] / (x^3 - 2)^2\n\nThen just multiply the numerator and combine like terms to get:\n\n[ -x^4 - 21x^2 - 4x ] / (x^3 - 2)^2\n\nDid that help?  Let me know if you have additional questions!\n\n.