According to the definition of a derivative : lim (f(x+h)-f(x))/h, what is the derivative of this function?
92912
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.
801052
It seems you are asking for f'(2)?  in which case, we have\nf'(x) = 1/(1+x^4)so that f'(2) = 1/9\n\nSteve
754408
you are right, derivative is NEVER zero. In fact, it is always greater than zero. This means that the function is always increasing on the whole real line, minus the point x = -1 since the function is not defined there. there is no relative extrema since the function is monotone increasing and the interval are open intervals(no relative extrema at the starting and ending points of the interval)
506639
The first derivitive of f(x) = 5/x^3 - 3/x^1/3 is...\n\nf'(x) = -15/x^4 + 1/x^(4/3)\n\nHope this helps. I used the product rule. Remember, you can write the term 5/x^3 and 5*x^-3. I find it simpler to rearrange the equation and use the product rule to get at the first derivitive.
679340
By definition, you would do the following:\n\nf'(x) = lim(h->0) (f(x+h) - f(x))/h\nf'(x) = lim(h->0) ((3(x+h) - 3x)/h\nf'(x) = lim(h->0) 3h/h\nf'(x) = lim(h->0) 3\nf'(x) = 3\n\nOr, you could simply do use the following fomulas:\n\nd/dx[af(x)] = a(d/dx[f(x)])\nd/dx[x^n] = nx^(n-1)\n\nd/dx[3x] = 3(d/dx[x]) = 3(d/dx[x^1]) = 3*(1*x^0) = 3\nf'(x) = 3
34808
F'(x) is the derivitive of F(x), which means it describes the change in F with respect to x.  Another way to say this is that F'(x) is the slope of F(x).  \n\nTo sketch the graph of F'(x), look at your graph of F(x).  Pick a point on that graph (maybe the first point). See what the slope (remember rise over run?) of F(x) is at that point.  That slope is the derivitive F' for that value of x.  Plot this on your F'(x) skecth.  Go the the next point and repeat.
957360
The symmetric difference quotient is\n\n[f(x+h) - f(x-h)]/2h.\n\nIt is the average of the right and left quotients [f(x+h) - f(x)]/h and [f(x) - f(x-h)]/h, where h > 0. So it tends to give a better approximation of the derivative.
1005009
A derivative is a function that provides the limit of the slope of a section of a primary function as the length of that section approaches zero.
3246
So for the first derivative:\n\n4(x-1)\n-----------\n3x^(2/3)\n\nThe second derivative:\n\n4(x+2)\n-----------\n9x^(5/3)
727467
It can be a constant or a variable.  It asks what is the behavior of the function as x approaches the "a" variable or constant.  Thus,\nlim x->a [(x - a)²]/[1 + cos (π/a x)]\n\nLet me show you.  We first get the limit:\n_______________________________________\n| Since it is the indeterminate form 0/0, we can use the \n| L'Hopital's rule.\n| lim x->a [2x - 2a]/[- π/a sin (π/a x)]\n|\n| Again, it is 0/0, so use L'Hopital's Rule,\n| lim x->a 2/(-π²/a² cos (π/a x)]\n|\n| We can now substitute x\n| = 2/(-π²/a² cos π)\n| = 2/[-π²/a² (-1)]\n| = 2/(π²/a²)\n| = 2a²/π²\n|__________________________________\n\nWe now have the limit\n= 2a²/π²\n\nNow, you can interpret this limit in 2 ways.\n\nIf "a" is a constant, then the limit of that function (of only 1 variable, x) when the variable x approaches an unstated constant "a" is the limit that we have acquired (which, in this case, is a constant).\n\nNow, if "a" is a variable, then the limit of that function (of 2 variables, a and x) when the variable x approaches the value of the variable a, is a function of a single variable, which is the limit we have acquired (which, in this case, is a function).\n\n^_^
633611
You need to check the glossary of the text you are using.\n\nSome writers invent their own symbols. In this case, I would guess, from your remarks, that the symbol means "the first derivative of the function f(x)".\n\nIn some calculus texts, the symbol used is f '(x),\n\n read "f -prime of x, or "the first derivative of f(x)".\n\nHope that helps!
1121843
the given function is not differentiable at x=0\n\ntherefore,   f(x) = x   for x>0        dy/dx = 1\n                           f(x) = -x  for x<0        dy/dx = -1
909269
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.
887496
The first condition for a function to be differentiable in a point x is that the function is continuous at that point.\nit's easy to prove the function is continuous at x=1\nbecause:\nLim f(x)=Lim(x^2 + 1)= 2 for x going to 1 from left (1-)\nLim f(x)=Lim(2x)=2 for x going to 1 from right (1+)\nWe have now proved that the function is continuous at x=1\n\nWe will look at f' at x=1\nf'(x)=2x ; x<=1\nf'(x)=2 ; x>1\nthe function is differentiable in a point x=a if and only if, the function is continuous and f' exist and has the same value when x goes to point x=a from left and right\n\nLim f'(x)=2 for x going to x=1 from left (1-)\nLim f'(x)=2 for x going to x=1 from right (1+)\nnow we have proven that the function is differentiable at x=1
921060
In general if you have x^n, the derivative is n * x^(n-1).\n\nFor x² the answer is 2x^1 or just 2x.\n\ny = x²\ndy/dx = 2x
1047841
By the def\n\nlim f(x + h ) - f(x)\n              /h\nPlug in (x+h) into f\nlim (x+h)^2 + 5(x +h) - (x^2 + 5x)\n                        /h\nFoil\nlim x^2 + 2xh + h^2 + 5x + 5h - x^2 - 5x\n                          /h\n\nCancel like terms\nlim 2xh + h^2 + 5h\n           /h\n\n\nReduce everything by h\nlim 2x + h + 5\n\nTake the limit and\nPlug in 0 for h\n2x + 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)