The second derivative?
407062
Just as the first derivative is the gradient of a graph, the second derivative is the gradient of the gradient.  It tells you how the slope is changing.  So when the slope is zero and the slope is getting more negative (turning down) then the stationary point must be a maximum.  If the second derivative is positive at that point, then the slope is turning up and the point must be a minimum.  If the second derivitive is 0 at that point, then it is an inflection point (slope going from positive to negative or negative to positive).  Hope that helps.
1069777
differentiate both sides:\ncos(x^2/y^2) * derivative(x^2/y^2) = 1\ncos(x^2/y^2) * (y^2 * 2*x -  x^2*2*y*(dy/dx))/y^4 = 1\ncos(x^2/y^2) * 2xy(y-x*(dy/dx))/y^4 = 1\ny-x*(dy/dx) = y^4/(2*x*y*cos(x^2/y^2))\n(y - y^4/(2*x*y*cos(x^2/y^2)))/x = dy/dx\nAssuming I didn't make any errors.
948927
well, to differntiate a funtion using first principle, you have to follow this procedure. \nlet f(x) be a function. then f'(x) is the derivative of f(x) with respect to x. from the first principle, it is given by\nf'(x)=  Lt        f(x+h)-f(x)\n        (h->0)    -------------\n                          h\nwhere Lt represents the limit as h tends to zero.\nnow, take f(x)=cos(x^1/2). so, f'(x) is\n         Lt        cos((x+h)^1/2) - cos(x^1/2)\n        (h->0)    -------------------------------\n                                    h\n    here we apply L'Hopital's rule, because substituting h=0 will yield an indeterminate form. so, the result is \n      Lt        -sin((x+h)^1/2)\n     (h->0)   ------------------\n                   2*((x+h)^1/2)\nsubstitute h=0, we get the result as -1/2*(sin(x^1/2)/(x^1/2))
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.
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.
438946
A way to estimate a derivative and define a general way of solving differential equations numerically. See the links for more information.
421380
from first equation put the value of b in second and then form a second degree differential equation in "a"...good luck!
832910
yes the integral of the derivative of f(x)=x^2 is x^2.\n\nand just to check:\nf(x)=x^2\nf'(x)=2x\nF(x)=2*(1/2)x^(1+1)=x^2\n\nwhether you graph it or not makes no difference.
977007
Definitely not.  Just do the derivative the way you normally would...\n\nRecall, for f(x)=x^n, f'(x) = nx^(n-1), that's how to handle the exponent\n\nuse chain rule...\n\nfor f(g(x)),   f'(g(x)) = f'(g(x))*g'(x), remember...
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
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.
92912
I'm on a roll working with the definition, so I\nthought I'd try this one too. \n\nf(x)=4x/(5x+2)\n\nSubstitute it into the derivative definition:\n\nf'(x)=lim h->0 ((4(x+h)/(5(x+h)+2)) - 4x/(5x+2))/h\n\nMultiply the adjacent terms:\n\nf'(x)=lim h->0 ((4x+4h)/(5x+5h+2) - (4x/(5x+2)))/h\n\nCombine the fractions:\n\nf'(x)=lim h->0 (((4x+4h)(5x+2)-(4x)(5x+5h+2))/((5x+5h+2)(5x+2)))/h\n\nMultiply the adjacent terms:\n\nf'(x)=lim h->0 ((20x^2+20xh+8x+8h-20x^2-20xh-8x)/(25x^2+10x+25xh+10h+10x+4))/h\n\nSimplify:\n\nf'(x)=lim h->0 (8h/25x^2+20x+25xh+10h+4)/h\n\nCancel out the "h" from the top and bottom:\n\nf'(x)=lim h->0 8/25x^2+20x+25xh+10h+4\n\nSince h being zero won't cause an infinity, take the limit:\n\nf'(x) = 8/25x^2+20x+4\n\nBut notice that the denominator can be factored:\n\nf'(x) = 8/(5x+2)^2\n\nYou might notice that this equal to answer #1,\nexcept that it is reduced differently.
947549
Even 'tho it's obvious, you really should force yourself to say 'derivative with respect to x' because, later on, it won't be quite so obvious what the derivative is being taken with respect to ☺\n\nFirst of all √(4-x²) = (4-x²)^½.  So\n(d/dx)(4-x²)^½ \n½(4-x²)^(-½) (d/dx)(4-x²) (derivative of an expont)\n½(4-x²)^(-½) [(d/dx) 4 - (d/dx)x²] (derivative of difference)\n½(4-x²)^(-½) [0 - 2x] (derivative of constant and exponent)\n½(4-x²)^(-½) (-2x)  (algebra)\n-x(4-x²)^(-½)  (algebra)\n-x/(4-x²)^½ (algebra)\n1x/√(4-x²) (more algebra ☺)  which is the final answer.\n\nSee why you needed to learn algebra? ☺\n\n\nDoug
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.
754435
I | AUG is not correct.  You cannot differentiate power functions by the exponent rule.  (Even doing that, she made an error.)  The implicit differentiation done by Greg G is correct, or you can use the functional power rule:\n\nd(ƒ^g)/dx = (ƒ^g)*[(dƒ/dx)*(g/ƒ) + (dg/dx)*lnƒ]
745843
Use the quotient rule\nThe derivative is\n[(Q)(25000 +150 + 3Q)' - (25000 + 150 + 3Q)(Q)'] / (Q)^2\n= [(Q)(3) - (25000 + 150 + 3Q)(1)] / Q^2\n= [3Q - 25000 - 150 - 3Q] / Q^2\n= [-25000 - 150]/ Q^2\n= -25150/Q^2\nIf you want (and I would recommend this), then you can combine 25000 and 150 to get 25150.
965952
Implicit differentiation is really jsut an exercise in using the chain rule. For instance, if you had e^(x^2), its derivative would be e^(x^2) * 2x, since d(x^2)/dx = 2x. It's the same thing with implicit, except that you don't actually know what dy/dx is yet, so you just write dy/dx. So for my example, d(e^y)/dx = e^y dy/dx. Now, onto your problem:\n\nd(x/(x+y))/dx = ((x+y) - (1+dy/dx)x)/(x+y)², or being simplified, (y-x dy/dx)/(x+y)²\n\nOf course, this assumes you are differentiating with respect to x. If you are differentiating w.r.t. a third variable (e.g. t), then you would have to use the chain rule on x as well. So your derivative would be:\n\nd(x/(x+y))/dt = (dx/dt (x+y) - (dx/dt + dy/dt)x)/(x+y)²