need help with a derivative?
967733
OK, its a product of two functions, where the rule is:\n\n(fg)' = f'g + fg'\n\n( (sin2 t+2) cos t )' = (sin2 t+2)' cos t + (sin2 t+2) (cos t)'\n\n(cos t)' = -sin t\n\n(sin2 t+2)' is a sum of two functions where you use (f+g)' = f' + g'\n\ntherefore (sin2 t+2)' = (sin2 t)'+2' = (sin2 t)' because 2'=0.\n\n(sin2 t)' is a composite function (first you calculate the sin of t and then you square the result), where you use: f(g(t))' = g(t)' * f'(g(t))\n\nSo it is 2*sin(t)*cos(t).\n\nSo the result is:\n\n2*sin(t)*cos2(t) - sin(t)*(sin2 t+2)
773933
1-In general the derivative of ln|U| is U'/U. In your case U = 2x^x + 4x + 1. Using the fact that x^x = exp(xlnx) you can easily check that the derivative of \n\nU' = 2(1 + lnx)x^x + 4.\n\nThus the derivative of your function is\n\n[2(1+lnx)x^x+4]/[2x^x+4x+1].\n\n2- The derivative of U^a, where a is a real number, is aU^{a-1}U'. In your case U = x e^x and a=1/2. Since U' =(x+1)e^x, one has\n\n(sqrt(xe^x))' =1/2{(x+1)/sqrt(xe^x)}e^x = 1/2{(x+1)/sqrt(x)}exp(x/2).
702852
equation of the tangent of the line is the derivative of the line\n\ndy/dx = 3x^2 + 2
43939
is the above question in first degreee derivative??
945648
Take the derivative of f :\nf'(x) = 5x^4 (x+7)^4 + 4 (x+7)^3 x^5\nf'(x)= x^4 (x+7)^3 [ 5 (x+7) + 4x]\nf'(x)= x^4 (x+7)^3 [ 9x + 35]\n\nSo the derivative is equal to 0 for the values :\nx= -7\nx= - 35/ 9\nx = 0\n\nthe function is increasing in [-11;-7],\nthe function is increasing in [-35/9;+12],
669809
Start by finding the partial derivatives.
517319
Using the definition of the derivative.\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n\nLet f(x) = mx + b\nf(x+h) = m(x+h) + b\n\nf(x+h) - f(x)\n= m(x+h) + b - (mx+b)\n= mx + mh + b - mx - b\n= mh\n\n[f(x+h)-f(x)]/h \n= mh/h\n= m\nTherefore,\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n= lim(as h->0) of m\n= m
87883
First write y as a product instead of a quotient:\n\n(2+x)*(2-3x)^(-1)\n\nNow you should be able to do this. \nUse 2+x as your f and (2-3x)^(-1) as your g. \nYou will have to use the chain rule to find g'.
1016838
Use the quotient rule for differentiation.  Taking the case where the firetruck is approaching:\n\nThe numerator of the derivative has the denominator of the original function times the derivative of the numerator minus the numerator of the original function times the derivative of the denominator.  \n\nThis is pretty easy, since the numerator of the original function is a constant.  The derivative of a constant is 0.  When approaching, you only have to worry about the original numerator times the derivative of (331 - v).  The 331 is a constant with a derivative of 0 leaving you the derivative of -v.  That's -1.  \n\nTherefore, the numerator of your derivative is -132400 * -1, or 132400 when approaching.\n\nThe denominator of the derivative is just the square of the original denominator, or (331 - v)^2 when approaching.  When approaching, that would give you 301^2 (301 squared), or 90601. \n\nWhen approaching, the rate of change would be 132,400/(301^2), or about +1.46.\n\nFinding the rate of change when moving away works the same, but with a +v instead of a -v.
739947
To do this you find the first derivative:\nf'(x) = 3x^2 -8x + 3\nWhen f'(x) = 0 there is a miinimum or maximum.\nWhere is this one increasing to a max and/or min?  Set it equal to 0,\n3x^2 -8x + 3 = 0; apply the quadratic formula to find the 0's for x:\n\nx = {-(-8) +- sqr[(-8)^2 - 4(3)(3)]}/2(3)\n   = {8 +- sqr[64-36]}/2 = {8 +- sqr[28]}/2 = {8 +- 2sqr(7)}/6\n   = (4+-sqr(7))/3 = x1, x2\n\nNow you have two values of x; now look at f(x) in the interval\n(-infinity, x1), then (x1,x2), then (x2,+infinity); this will tell you the answer.  Make a table showing the values approaching x1, \nx1 to x2, then x2 on.  And there you have it.
713982
use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
844269
assuming that your question is V(w)=w^2*(3/(8w^3)),\nwe can simplify it as V(w)=3w^2/8w^3 = 3/8w\nso, the derivative will be V'(w) = -1*3/8w^2 = -3/8w^2
758976
The total differential is [∂F/∂x]dx +[ ∂F/∂y]dy\n\n∂F/∂x = 4x^3*y^-1\n∂F/∂y = x^4*(-y^-2) = -x^4y^-2
176854
You can graph it but probably that is not the answer your teacher is looking for. If you wanna do this by the good old method:\n\nf'(x)=3^x*ln(3)\n\nYou are looking for a point where the slope equals to 1 meaning 1 equals to your derivative.\n\n3^x*ln(3)=1\n\nNow use the calculator: x=-0.0856
816542
Find derivatives with respect to each variable.
934742
uh, find the derivative mathematically and then graph it.\n\nfor example to graph derivative of x^2.\n\nfind the derivative equation f(x)=2x\nand then graph y=2x.\n\nooops, didnt realize it was a picture.\n\nfind the zeroes (a high or low point i.e. tangent is a horizontal line)\nyou should be able to track poitive and negative slopes.  not too hard, you'll figure it out.
761290
well first you have to get the first derivative\n\nso: the 1st derivative= t*cost+1*sint\n                                  = cost^2+sint\nthen you take the derivative of that\nso: the 2nd derivative= 2cost*-sint+cost\n                                   = -2sintcost+cost