Find y^n in the following( second derivative): y=sin(x^2.e^x)?
967981
y'=cos(x^2e^x)(2xe^x+x^2e^x)\n\ny''=-sin(x^2e^x)(2xe^x+x^2e^x)^2+\n      (2(e^x+xe^x)+(2xe^x+x^2e^x))cos(x^2e^x)
178683
yeah, I suppose you want me to factorise x^n - y^n. the general formula is\n\nx^n - y^n = (x-y)(x^n-1.y + x^n-2.y^2 + x^n3.y^3+.........+x^2.y^n-2 + x^n-1.y^1)\n\nTry putting n=2 or 3 or4 in this expression and you'll see what I mean.\n\nYou'll arrive at nothing but\n\nx^2 - y^2 = (x-y)(x+y)\nand so on..\n\nAnd please, the next time, a question without exponents. i'm SICK of '^'. \n\nAnd you know what, mr. 'timmobilelaptop' , the next time try using your brain. It's better than any T-189 calculator.
764897
use derivative rule for exponential as\n\nd/dx(A exp(bx))=(A*b)*exp(bx)\n\nyou got hint. No more Spoon Feeding
1049778
y=f(x)=((x-5)/(2x+1))^3\nIf y=f(x)^n then    :    y'=n*f(x)^(n-1)*f'(x)\nNow here:\ny' = 3 * ((x-5)/(2x+1))^2 * [(1*(2x+1)-2*(x-5)) / (2x+1)^2]\nSimplifying:\n\ny'=33*(x-5)^2/(2x+1)^4
844269
assuming that your question is V(w)=w^2*(3/(8w^3)),\nwe can simplify it as V(w)=3w^2/8w^3 = 3/8w\nso, the derivative will be V'(w) = -1*3/8w^2 = -3/8w^2
1181156
It looks like that first 1 should be an i. In this case,\ni^z=exp(i*pi/2*z),\nwhere I am using the principal branch of the logarithm to define the exponential.\nThen the derivative will be \n(i*pi/2)*exp(i*pi/2*z)=\npi/2*i^(z+1)\n\nEasy.
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...
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)
982407
dy/dx=cosx*e^y*e^sinx\ndy/e^y= cosx*e^sinx*dx\ne^-ydy= cosx*e^sinx*dx\nintegrate\ne^-y/(-1)=int. cosx*e^sinx*dx\nput sinx=t\ncosxdx=dt\n-e^-y=int.e^tdt\n        =e^t+c\n        =e^sinx+c\nput x=0,y=0\n-1=1+c\nc=-2\n-e^-y=e^sinx-2
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).
669001
e^f(x) = x is the best answer i can give you.
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.
26362
Part 1:\n\nWe can rewrite y^x as (e^ln(y))^x which is e^(xln(y))\n\n∂f/∂x f(x,y)= ∂f/∂x (x^y) + ∂f/∂x(y^x)\n\n∂f/∂x f(x,y)= yx^(y-1) + ln(y)e^(xln(y))\n\n∂f/∂x f(x,y)= yx^(y-1) + ln(y)y^x\n\n\nPart 2: There is no simple function which is the integral of e^(x^2).
805634
It is easier (by far) to do this with logarithmic differentiation:\ndy/dx=y*d/dx[ln y].\n\nHere, with y equal to your expression,\nln y=ln 10 +ln x +4ln(x-1)\n-3ln(x-2)-2ln(x+1),\nso the derivative of the logarithm is\n1/x +4/(x-1) -3/(x-2)-2/(x+1).\nThe first derivative is now your original expression times this.\n\nTo find the second derivative, use the product rule. The derivative of y is already known, and the second derivative of the logarithm is easy.
979217
y = sqrt(1+t^2) = (1+t^2)^(1/2)\n\nby the power and chain rule\n\ndy/dt = (1/2)(1+t^2)^(-1/2) x d/dt (1+ t^2)\n\n= 2t(1/2) /sqrt(1+t^2)\n\n= t/sqrt(1+t^2)\n\nTherefore\n\ndy = (t / sqrt(1+t^2)) dt
276671
(e^(cose^e+sin^e)/2)^pi\nhahaha ..just kidding
931612
f^{-1}(y) = (y-42) / 7\ns