derivitive problems, chain rule?
831528
"What i know: it can also be written as (x^2 + cos x) ^ -1/2\nand the derivitive of that = (2x - sin x) ^ -1/2"\n\nWhat you said here was to take the derivative of the inside function, but what you do is take the derivative of the outside function and multiply it by the derivative of the inside function.\n\nIf you have a composition of two functions such as h(x) = f(g(x)), then h'(x) = f'(g(x)) * g'(x) [or as in your notes, f(g) * g']\n\nThat is why you must use the chain rule for 1 and 3 but not for 2, since 2 does not involve a composition:\n\n1. f(x) = (x^2 + cos x) ^ (-1/2)\n\nf(x) = g(h(x)) where g(x) = x^ -1/2 and h(x) = x^2 + cos x\n\n3.  f(x) = 2 ^ (x^2 - sin x)\n\nThis isn't a product- it's an exponential function of another function.\n\nf(x) = g(h(x)) where g(x) = 2 ^ x and h(x) = x^2 - sin x
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2
761104
The first derivative is correct. The second derivative is:\n\n- [ln 6x+1] / [ (xln(6x))^2 * ln(ln(6x))] - 1 / [ln(ln(6x)) * (xln(6x)) ]^2\n\nEssentially you want to use the product rule considering the first derivative to be:\n\n1/xln(6x) * 1/ln(ln(6x))\n\nd[1/xln(6x)]/dx = -1 * d[xln(6x)]/dx * [xln(6x)]^(-2)\n                        = -1 * [ln(6x)+1] / [xln(6x)^2]\n\nd[1/ln(ln(6x))]/dx = -1 * d[ln(ln(6x))]/dx * [ln(ln(6x))]^(-2)\n                           = -1 * [1 / xln(6x)] / [ln(ln(6x))]^2\n\nAnd then you use the product rule: (u * v)' = u' * v + u * v' to end up with what I gave you above.
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
347106
Okay, if you're actually serious, you REALLY should be paying more attention in class, but I'm gonna answer it for two points anyway.\n\nThe derivative of x^2 (with respect to x) is 2x.\n\nHave fun.
892157
wait 1 sec.\nOk this is reasonably difficult, you have to use the chain rule and the quotient rule.\nUsing the chain rule:\nderivative = [ (1/2) * ( (2x + 3) / (7x+4) ) ^ (-1/2) ]  *  [d/dx (2x+3) / (7x+4) ]\n\nSo now you to use the quotient rule to find:\nd/dx (2x+3) / (7x+4) = (14x+8 - 14x + 21) / (7x+4)^2\n= 29 / (7x+4)^2\n\nSo the derivative is [1/2 * ((2x+3)/(7x+4))^(-1/2)]*[29 / (7x+4)^2]\n\n\nBy the way i got the thumbs down before I wrote all this up, so it doesn't mean it's wrong.
981618
I guess 1) just wants you to write: Want dA/dt, have da/dt and db/dt.\n(From now on I'll use A' to mean dA/dt etc).\n2)\nA=π*a*b, so using the product rule, A' = πa'b + πab'. So A' = 7πb - 11πa.\n3) Just substitute in: A' = 7*π*50 - 11*π*30 = 20π, so thats the rate it is increasing.\n4) Again substitute in, A' = 7*π*20 - 11*π*40 = -300π, so now its decreasing at that rate.
766600
You have to use the quotient rule for this problem:\n\nFor any function f(x) defined by, f(x) = g(x)/h(x),\n\nf'(x) = [g'(x)h(x) - h'(x)g(x)]/[h^2(x)]\n\nIn this problem, we have:\n\ng(x) = xe^-x\nh(x) = x^3+x\n\nTherefore, \n\ng'(x) = xe^-x(-1) + e^-x(1) [you have to use product and chain rule for thie derivative]\n\nSimplfying, you get e^-x(1-x)\n\nh'(x) = 3x^2 + 1 (power rule)\n\nNow that we have all parts of the derivative, just plug in:\n\nf'(x) = [e^-x(1-x)(x^3+x) - (3x^2+1)(xe^-x)]/(x^3+x)^2\n\nNow just simplify this ugly mess =P\n\n---------\n\nHope this helps
183101
You can use the product rule the way it is written right now, or you can use foil and then take the derivative the regular way.  I think the second would be less work.\n(x+1)(x+3)\n= x^2 + 4x + 3\nThe derivative is 2x + 4\n\nIf you use the product rule, you would have\n(x+1)(x+3)' + (x+1)'(x+3)\n= (x+1)(1) + (1)(x+3)\n= x+1 + x+3\n= 2x + 4
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.
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)
372419
I fear you have a missing element here... do you want the differencial of 4P^2..etc because if so you need to know what 4P^2..etc equals, or is 4P^2..etc the differencial of another expression, because if so your question is a bit like saying what is the answer to x=2\n\nIf you want to differencial of 4P^2..etc then I suggest you solve the original equation for t which will make the algebra easier and then use the chain rule where dP/dt = [1/(dt/dP)]
786389
The derivative is [(x-4)- (x+1)]/(x-4)^2 = -5/(x-4)^2\n\nIf we use the definition of the derivative:\n\n\ndy/dx = Lim h -> 0 {f(x+h) -f(x)}/h\n= lim h-> 0 [(x+h+1)/(x+h-4) - (x+1)/(x-4)]/h\n                   (x+h+1)(x-4)-(x+1)(x+h-4)\n= lim x-> 0 ---------------------------------------\n                   (x+h-4)(x-4)h\n\n= lim h-> 0 -5h/(x+h-4)(x-4)h = -5/(x-4)^2
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
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!
904043
One variable is related to another by a function involving constant terms and terms of 1st order or higher.
79944
The above is the derivative.\nIntegration gives x + (1/4)x^4 + c (generic constant).