Differentiation question...?
349088
Since A is a constant, you can take it outside and multiply by A at the end.
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.
754733
Let y = u/v   where u=some function and v =some funtion\n\nthen dy/dx = (v*du/dx - u*dv/dx)/ (v^2)     **Quotient Rule**\n\nIn this case u = 6x^2-1 and v = 2*sqrt(x)\n\nSo you just differentiate and plug into the equation for dy/dx
888322
The equation is continuous at both points but is not differentiable at either of them.\n\nYou have f(x)= Ix^2+5x+6I=|(x+2)(x+3)|=|x+2| |x+3|\n\nNear x=-3 |x+2| is continuous and differentiable because x+2 is strictly negative. The problem is that near x=-3 |x+3| is not differentiable. You basically follow the argument that you gave in your problem statement. So you have something like this:\n\nf(x)=g(x)h(x)\n\nwhere g(x)=|x+2| and h(x)=|x+3|. Using exactly the argument that you provide h(x) and g(x) are continuous at x=-3. However, h(x) is not differentiable while g(x) is so the product is not differentiable.\n\nFor the case x=-2 you basically use the exact same reasoning for this new point. Again, you express f(x) as the product of g(x) and h(x) and then use the same argument that you provide in your problem statement.
739543
To differentiate y= 2/x\n\nFirstly you get rid of the fraction by taking X to the top(this will make its power negative) giving:\n\ny=2x^-1\n\nFrom here its just a simple case of carrying out the differentiation:\n\ndy/dx = -2X^-2 or dy/dx = -2/X^2
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
706213
Wait a minute, did you want to know when the function is zero or when its derivative is zero?  \n\nFor derivative:\ny' = 2x + sin x, right?\n\nIt looks like there's one place where it's 0.  Requires and numerical solution.\n\nWell, it's 0 at x=0, right?\nI assume that x is in radians.  1 radian is about 57 deg.  As x decreases from 0 both parts become more negative.\n\nIt looks like x = 0 is the only zero of the dirivative.
816980
1 is false. The function is continuous.\n2 is false; see above.\n3 is true.  Hence, 4 is false, and so is 5.\nYou are right about the non-differentiability at x=3.  A function cannot be called differentiable if its slope at x- does not approach its slope at x+ as the interval approaches zero.
840388
This is differentiation of the composite function \n\nJ(x) = I(a(x),b(x))f(t)dt\n\nwhere \nI(a,b) is the integral from a to b, \na(x) = tan(x), b(x) = x^2 \nand f(t) = sqrt(1+t^4).\n\nApply the Chain Rule and then the Fund. Thm of Calc., which implies that dI/da = -f(a) and dI/db = f(b). These should be written with curly d's because they are partial derivatives of I (as a fctn of the two integration limits). And work out da/dx and db/dx. So \n\ndJ/dx = (dI/da) (da/dx) +  (dI/db) (db/dx)\n = -f(a(x)) (da(x)/dx) + f(b(x)) (db(x)/dx)\n\nNow just put in your particular functions f, a and b.
1116918
In Calculus of Varations deals with minimization problems the most elementary of which is the Brachistochrone.  This is the problem of the shape of a wire which provides the fastest path between two points for a marble sliding down it.  I don't know what applications of differentiation is.
1107625
I assume you are able to take partial derivatives. So take the partial derivatives that appear in the heat equation (you did look up the heat equation, right?) and plug them in to see if you get equality.\n\nSo, the partial derivative with respect to t is\n-lambda^2 k*u(x,t). This is the same as k times the second partial derivative with respect to x. Looks right to me!
277316
let |f|=k (some constant).let u,v be the real & img parts of 'f'.\n            the we have |f|=u^2+v^2, u^2+v^2=k^2\n      differentiating partially,wrt to x&y ,and \n\n    USING CAUCHY-REI EQ'S FROM TWO RESULTING EQ'S \n\nu finally get     (u^2+v^2)(du/dx)=0.\n                 &  (u^2+v^2)(du/dy)=0. \nfrom above two eq's   we have \n                     k^2=u^2+v^2=0  implies u=0&v=0\n\n                  so 'f' is constant.
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
956017
well, to differentiate a the given funtion with respect to                (1-4x^2)^1/2, we assume (1-4x^2)^1/2=y. and the function becomes sin^-1(4y) or 1/sin4y. we have to differentiate this with respect to y.\nso, the function, which is cosec4y, on differentiation gives the result as -cosec4y*cot4y*4\nsubstituting the value of y we get the final result in terms of x as\n-4cosec((1-4x^2)^1/2)*cot((1-4x^2)^1/2)\nthe long string in the cot is (1-4x^2)^1/2 ie., y
931778
Since you have conflicting answers, I'm just chiming in to say go with ManiacalAce -- he got the correct answer.\n\nCorrect second derivative is as follows:\n\n(-9/4)x^(-3/2) - (3/4)x^(-1/2)\n\n~ ♥ ~
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
434625
You are trying to find d/dt (0.51) ...\n\n0.51 -> alpha [ which inturn points "CONSTANT" ]\n\nDerivative of a constant is zero\n\nso d/dt (alpha) = 0