derivative of trigonometric function?
916561
This is a fraction, yes?\n\nThe derivative of a fraction is\n\nThe derivative of the numerator times the denominator\nminus\nThe derivative of the denominator times the numerator\nAll divided by the square of the denominator\n\nLet's do it:\n\n[-sin(x) * (1 - cos(x) - sin(x) * (1 + cos(x)] / [1 - cos(x)]^2\n[-2sin(x)] / [1 - cos(x)]^2\n\nNotice that your function is discontinuous at x = 0 and x = pi, etc., and so is the derivative.\n\nI LUV screwed up again. The derivative of cos(x) is -sin(x), not sin(x) as she has it.\n\nSteve's solution agrees with mine.
672016
Recall that the derivative function of the function f(x) = arctan(x) is the function f'(x) = 1/(1 + x^2). Consider the derivative function of the related function g(x) = arctan(a/x) where a is some real constant. Use the chain rule and you will find that it is only a constant multiple away from your integrand.
184556
Let x(t) = A*cos( w*t + theta )\n\nthen x(0) = A*cos( theta )\nand dx/dt (0) = -A*w*sin(theta)\n\nso if you know x and its derivative in t=0 u have two equations and two unknowns ( A and theta ) . the rest is simple algebra .
742056
Basically, you'll have to split it up into cases. If you draw the graph of |sin x|, you'll notice that theres a 'spike' everywhere it hits y=0, so its not actually defined there. However, in the other cases, you just take the normal derivative, but make it positive.\nSo you get something like:\nIf x is a multiple of pi, its undefined.\nIf sin x > 0, then the derivative is cos x. (ie for x from 0 to pi, 2pi to 3pi, 5pi to 7pi..)\nIf sin x < 0, then the derivative is -cos x. (ie for x from pi to 2pi, 3pi to 4pi..)\n\n(Note thats not the same as |cos x| - the derivative is definitely negative at some points).
1048691
(r*x)*(-sin(x))+(cos(x))*(r)+\n(0)*(lnrsin(x))+(r)*(cot(x)) \n\nfor the cot(x) you end up for the derivative of lnrsin(x) is (1/sin(x)) *cos(x)=cot(x)\n\nsorry it cut off on the top. hopefully this works
741912
cos ^1 t = x such that cos x = t\n\nsin ^1 t = y such that sin y = t\n\nThese are called inverse trigonometric functions.\n\nnow cos^-1 t = sin^-1  t = x \nso  cos x = sin x = t\n\ntan x = 1\n\nx =  pi/4\n\nthe above is solution
72362
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
766972
First off, this problem involves the chain rule. I think the person who said package rule really meant product rule, which is not requried here. \n\nThe chain rule allows us to find the deriviative of a function which we don't know how to differentiate in terms of functions which we do know how to differentiate. In this case, we know how to differentiate sin(3x), but not (sin(3x))^4 (I think it is less confusing to write it that way given the lack of mathematical notation available here).  The chain rule, in general, says that, if we have two functions, g(x) and h(x), we can find the deriviative of the compostie function g(h(x)) as follows:\n(g(h(x))' = g'(h(x)) * h'(x)\nSo, let us have hour h(x) function defined as:\nh(x) = sin(3x)\nIt's derivative is then:\nh'(x) = 3cos(3x)\nNow, lets define g(x) as:\ng(x) = x^4\ng'(x) = 4x^3\ng composed on h is then:\ng(h(x)) = g(sin(3x)) = (sin(3x))^4\ng' composed on h is:\ng'(h(x)) = 4(sin(3x))^3\nNow we have everything we need to use the chain rule, and find the derivative of g(h(x))\n(g(h(x))' = 4(sin(3x))^3 * 3cos(3x) \n             = 12*cos(3x)*(sin(3x))^3\nIn practice, one doesn't need to do this process so formally. Any time you have some base function you know how to differentiate raised  to some power, to take the derivative you simply bring the power out front, lower the power by one, and mulitply by the derivative of the base.
764433
Finding max's and mins is done thru the first derivitive and setting it to 0, so\n\nf(x)= sinx - cosx\nf1(x)= cosx + sinx\n0=cosx + sinx\n-cosx = sin x\nthis occurs when x= (3pi)/4\nmax is root 2, as was stated.\n\nThe answer that used to be directly above is wrong.
1298
derivative of 5(sec(x)^2) with respect to sec(x) = 10(sec(x))\n\nderivative of 5(sec(x)^2) with respect to x = 10 * derivative of sec(x) with respect to x
245265
The derivative of sinx is just cosx, but the derivative of cosx is -sinx.  Then the derivative of -sinx is -cosx, and the derivative of that is sinx.  So it actually takes four applications of the derivative to get back where you started rather than two.  \n\nIf you're comfortable with the definition of the a derivative you can work it out and see that that's true.  Otherwise, just draw the graphs and look at where the slopes are negative and where they're positive.\n\nHope that helps!\n\nPS.  When you're looking at the page your question is on, you can click "additional information," and add more there rather than making a new question.  Saves you five points. :-)
882357
The derivative of ln|f(x)| is 1/f(x) * d f(x) / dx;  this will be\n\n1/[sec5x + tan5x] * d(sec5x) / dx + d(tan5x) / dx\n\nd(sec5x) / dx = d(1/cos5x) / dx = -1/cos^2(5x) * -5sin5x = 5tanx*secx\n\nd(tan5x) / dx = 5sec^2(5x)\n\n(5tanxsecx + 5sec^2(5x)) / (sec5x + tan5x)\n\nFactor out 5sec5x to get 5sec5x * (tan5x + sec5x)/(sec5x + tan5x) = 5sec5x
1034892
What is the function? I assume it is \n\n(dy/dx is equal to 1 only if y=x+C whre C is a constant)\n\nTo prove use the definition of the derivative which is dy/dx as a limit of delta_x appraching of (delta_y/delta_x)\nwe have \ny2-y1=x2+C-x1-C then\ndelta_y=x2-X1\n\ndelta_y/delta_x=(x2-x1)/(x2-x1)=1 QED\n(the is no need to take a limit, however in must be used. I hav eommited it for lack of better representation)\n\nBest
861008
NO\n\nf'(x)=(-sinx)(e^cosx)\nf'(4.0)=(-.07)(2.7)= -1.89
79944
The above is the derivative.\nIntegration gives x + (1/4)x^4 + c (generic constant).
274298
Use implicit differentiation to find dy/dx. Standard differentiation rules apply, except when you differentiate a term with a "y" in it you must tack on a dy/dx, so:\n\n2x+8y(dy/dx)=0\ndy/dx = -x/4y\n\nThe tangent line intersects the ellipse at (x1,y1), so the slope of the tangent line will be -x1/(4*y1).\n\nWe can also compute the slope of the tangent line by plugging the two points (-5,0) and (x1,y1) into the standard slope formula m = delta(y)/delta(x) and get that the slope of the tangent line is y1/(x1+5). Since these expressions both describe the slope of the tangent line, they must be equal, so\n-x1/(4*y1)=y1/(x1+5)\nx1^2+5*x1=-4*y1^2\nx1^2+4*y1^2+5*x1=0\n\nSince (x1,y1) is a point on the ellipse, it must satisfy the equation of the ellipse, so x1^2+4*y1^2=5. Substituting this in to x1^2+4*y1^2+5*x1=0 gives us 5+5*x1=0, and x1=-1. At this point, there are actually two y1 values that will satisfy the equation of the ellipse, y1 could be either plus or minus 1 (get these by letting x=-1 in the original equation of the ellipse), and so there are actually two points on the ellipse where the tangent line passes through (-5,0), and those points are (-1,1) and (-1,-1).
49770
There is no closed form for the nth derivative of this function.  I symbolically calculated and factored the first 50 derivatives on MathCAD, and there was not a pattern.  Sorry, but you're out of luck....