does anyone know the derivative of a negative (x)?? its calculus?
152962
The way to think of this is that -x is really -1 x. \n\nSo if I asked the derivitive of 10x, you know it is 10. And if I asked for -3x you know it is -3. So what is it for -1 x? -1, of course!
437876
Take the derivative:\n\nf'(x) = 2x - 4\n\nSet the derivative equal to 0, and solve for x:\n\n0 = 2x - 4\nx = 2\n\nSub this value of x into the original equation to find the turning point:\n\nf(2) = (2)^2 - 4(2) + 1\nf(2) = -3\n\nSo, your turning point is (2, -3).\n\nAnd, since the x-squared term in the equation is positive, you know that this point is a minimum.\n\nDone and done.
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.
749429
x = -7\n\nMultiply two negatives and you get a positive
887032
This is a fundamental theorem of calculus problem.\n\nThe derivative of the integral of t * sqrt (t^3 + 2t) dt from -1 to x is\n\nx * sqrt (x^3 + 2x)
705841
When you have your variable in the exponent, you can't take derivates like you would for x^n.\n\nWhen you have a constant raised to x, you take the derivative like this:\n\nd/dx ( a^x ) = a^x ln a\n\nSo, if f(x) = 2^x, then f '(x) = 2^x ln2\n\nHope that helped!\n\n.
676902
Incomplete question. Are you looking for the anti-derivative of x, a, or b???\n\nput dx, da, or db at the end of the equation so we'll know.
921398
I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
195729
you dirty dirty man... making me derive... *dusts off brain* alright, give this a shot...\n\nstarting with the inner terms....\n\nderivative of x... 1\n\nderivative of 4sin(x)... 4cos(x)\n\nderivative of 2... 0\n\nderivative of the larger term 1/(2(4sin(x)+2)^(1/2))\n{term needs to be taken to the bottom of the fraction as the exponent is negative}\n\nCompiled we get... 4cos(x)/(2(4sin(x)+2)^(1/2))\n\ninserting the 0... we get... 4/(2(2)^(1/2))... or 2/2^(1/2)\n\nrationalize... multiply by 2^(1/2)/2^(1/2) and we end up with....\n\n2^(1/2)
1020565
The derivative of the function given is\n\ny'=cos(2x)*2 - 2*cos(x) = 2(2cos^2(x)-1)-2*cos(x) \n= 4cos^2(x)-2cos(x)-2 = 2(2cos^2(x)-cos(x)-1)\n=2(2cos(x)+1)(cos(x)-1)\n\nFor y'=0, either 2cos(x)+1=0 or cos(x)-1=0, from which we get\n\ncos(x)=-1/2 (which gives either x=2*pi/3 + 2*pi*k, k an integer or x=-2*pi/3 + 2*pi*k, k an integer)\n\nor \ncos(x)=1, so x=2*pi*k, k an integer.
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)
765615
differentiate implicitly:\n\n2x + 2yy' = 0\n\nand then some algebra\n\ny' = -(x/y)\n\nthen differentiate again:\n\ny'' = (-y - (-x)*y')/y^2 = (-y + xy')/y^2\n\nthen substitute in values:\n\ny' = -0/5 = 0\n\nand y'' = (-5 + 0*0)/(5^2) = =-5/25 = -1/5
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.
711470
Try here:/\n\nhttp://archives.math.utk.edu/visual.calculus/
695915
Break this down into three functions depending on the value of x:\n\nWhen x<-1 you have f(x) = (-x-1)(2-x) because (x+1) and (x-2) are negative.  This is the same as f(x) = (x+1)(x-2)\n\nWhen -1<=x<2 you have f(x) = (x+1)(2-x).\nWhen x>=2 then f(x) = (x+1)(x-2).\n\nSo f(x) = (x+1)(x-2) = x^2-x-2 outside -1<=x<2,\nand f(x) = -x^2+x+2 in -1<=x<2.\n\nf'(x) = 2x-1 outside -1<=x<2, and f'(x) = 1-2x in -1<=x<2.
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....
1128734
I'm going to make a wild guess that you are trying to check an answer you got with the answer in the back of the book. I'm also going to assume that you were supposed to take a derivative of\n(x^3/(a-x))^(1/2).\nIn that case, you got the wrong answer because you didn't do the chain rule. You still need to multiply what you have by the derivative of the thing inside, i.e. the derivative of x^3/(a-x).\n\nIf I'm making a bad assumption and this didn't come from a differentiation problem, please ignore my answer.