Derivative of e^(3x^(2)-7x)?
352749
By the chain rule:\n[e^(3x^(2)-7x)]*(6x-7)
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
842257
divide through\n\ny = [x^(3/2)]/3 - 2x^(1/2) -x^(-1/2)\n\nthen take the derivative\n\ndy/dx = [x^(1/2)]/2 - x^(-1/2) + [x^(-3/2)]/2\n\nthen put x^(3/2) in the denominator\n\ndy/dx = [1/x^(3/2)]*{ [x^2]/2 - x + 1/2 }\n\nwe take out the one half to find\n\ndy/dx = [1/2x^(3/2)] * [ x^2 - 2x + 1 ]\n\nwhich is the same as writing \n\ndy/dx = [1/2x^(3/2)] * (x - 1)^2\n\nor a = -1 and b = 2
157895
Not quite sure what you mean, but I think this is what you're asking for:\n\nif you have y = x^3 + 2x^2 - x + 5\n\nto differentiate that, let's rewrite the equation like this: \n\ny = x^3 + 2x^2 - x^1 + 5x^0\n\nNow, for each item in the equation, multiply that item by its exponent, and subtract 1 from the exponent, giving you:\n\ny' = 3*(x^2) + 2*(2x^1) - 1*(x^0) + 0*(5x^-1)\n\nwhich simplifies to:\n\ny' = 3x^2 + 4x - 1\n\nAnd there you have the derivative of a simple polynomial equation.
806346
using implicit differentiation:\nyou get 2y(dy/dx) - [y + x(dy/dx)] -3 = 0\nthen move 3 over and factor out (dy/dx):\n(2y-x)(dy/dx)=3 + y\ndy/dx=(3+y) / (2y-x)\nnow plug in your point (0,-1)\ndy/dx = (3+ -1)/(2(-1) - 0\ndy/dx = -1\nanswer = -1\n\n(dont forget to use your chain rule when you are implicitlly differentiating)\n\n#2\nsame thing make sure you sue multiplication rule\n[y^2 + 2xy(dx/dy)] - 2(dx/dy) + 12y^2(dx/dy) = 0\nthen move over the y^2 and factor out dx/dy\n(dx/dy)[2xy - 2 + 12y^2] = y^2\ndx/dy = y^2/ (2xy -2 +12y^2)\nPlug in 1 for y\ndx/dy = 1/(2x -2 + 12)\ndx/dy = 1/ (2x +10) or (2x + 10)^(-1)\n\nAnswer (leave in terms of x) = 1/ (2x +10) or (2x + 10)^(-1)\n\nI may have made mistakes so let me know
849348
There are two methods, Method 1 you could simplify (1-X^4)^7 then multiply through by X^3, but... you dont want to do that, especially on an exam. \n\nMethod 2 is substitution.\n\nLets make up an equation here that we can substitute into our original equation...\n\nSay U=(1-X^4)  so we have (X^3)U^7dx, understand?\n\nNow lets take the derivative of U....  so dU= -4x^3dx, and solving for dx = dU/(-4x^3)\n\nLets plug that into our equation for dx.\n\nwe now have (X^3)*U^7dU/(-4x^3)\n\nCanceling the X^3 were left with (-1/4)*U^7dU\n\nNow you can integrate that right!\n\n(1/8)*(-1/4)*U^8 \n\nfurther simplified is\n\n(-1/32)U^8\n\nNow just plug back in U=(1-X^4) and we get\n\n(-1/32)*(1-X^4)^8\n\nThat should be the right answer!
857727
According to the chain rule and to the rule for the derivative of a product, we have y' = (x^2 +3) e^(x^2+1) (2x) + (2x) e^(x^2 +1) = e^(x^2 +1) (2x^3 + 6x + 2x) = e^(x^2 +1) (2x^3 + 8x)
314068
y = x^e^(7x)\nln y = ln [x^e^(7x)]\nln y = e^(7x)*ln x\n(1/y)*y' = 7*e^(7x)*ln x + (1/x)*e^(7x)\n(1/y)*y' = e^(7x)*[7*ln x + (1/x)]\ny' = y*[e^(7x)*[7*ln x + (1/x)]]\ny' = x^e^(7x)*e^(7x)*[7*ln x + (1/x)]\n\nTo ingameswetrust:\nHow could y = x^e^(7x) be mistaken for a differential equation? There are no derivatives anywhere in the equation. The author clearly states, "find d/dx..." However, if you want to solve a differential equation, then solve the following:\ndy/dx = x^e^(7x)*e^(7x)*[7*ln x + (1/x)]\n\nThere is one answer that should be obvious even to the casual observer.
965203
y = (ln x²)^(2x+3) = e^((2x+3) ln ln x²)\nNow just ruthlessly apply the chain rule:\ndy/dx = e^((2x+3) ln ln x²) * (2 ln ln x² + (2x+3) * 1/ln x² * 1/x² * 2x)\nAnd simplify:\n(2 ln x)^(2x+3) * (2 ln (2 ln x) + (2x+3)/(x ln x))
255922
Im not sure if you need clarification on how to do this. Ill provide a few details on how to get these answers:\n\n1) ∫sinxdx = - cos(x) + C\n-This one is relatively straightforward, just think "what gives me sin(x) when I take its derivative"\n\n2)∫e3xdx = (Im assuming this means e^(3x))\n         = (e^(3x))/3 + C\n-Just think of the chain rule backwards. If you take the derivative of e^3x you'll have to multiply by 3, so when you take the derivative divide by 3.\n\n3)∫cos3xdx = (Im assuming this is (cos^3)x)\n           = ∫((cos^2)x )(cos(x))\n           = ∫(1-(sin^2)x)((cos(x))\n      Now let u=sin(x)\n             du=cos(x)dx\n           = ∫(1-u^2)du\n           = [u - (u^3)/3]\n     Sub back in for u\n           = sin(x) - (sin^3(x))/3 + C\n-Wow, that one was hard to type out. I hope you can follow it somewhat. The trick is to write cos^3(x) as cos^2(x) and cos(x). Then the identity cos^2(x) + sin^2(x)=1 is used. The u substution allows us to find the integral at the end.\n\n4) Derivative of 3x^4 + 2x^2 + x + 3\n                  = 12x^3 + 4x + 1\n-The derivative is defined by some long equation, but you can use the following fact:\nIf f(x)=x^n\nthen f'(x)=nx^n-1\n\nHopefully you understand most of what I did. Im sure someone else will probably explain a lot of this in much better detail. Like I said, I wasnt sure how much detail you were looking for. \n\nPS Ill tell you site that will help you forever in any integral calculus course: http://integrals.wolfram.com/index.jsp
931612
f^{-1}(y) = (y-42) / 7\ns
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
761317
We have to find derivative of, y = the integral where lower limit is e^x and upper limit is 0 of sin^3 t dt.\n\ny = int(e^x)(0) [sin^3 t] dt     [I would write lower limit in first bracket and upper limit in bracket after that, so dont confuse]\n\nNow, the derivative of the integrand f(t)=sin^3 t is f'(t)=3 (sin^2 t) *cos t\ndy /dt= d {int(e^x)(0) [sin^3 t] dt }/dt\n=>dy /dt= d {f(t)}/dt (at 0) - d {f(t)}/dt (at e^x)\n=>dy /dt= f'(t) (at 0) - f(t) (at e^x)\n=>dy /dt= [3 (sin^2 t) *cos t] (at 0) - [3 (sin^2 t) *cos t ] (at e^x)\n=>dy /dt= [3 (sin^2 0) *cos 0] - [3 (sin^2 (e^x)) *cos (e^x) ] \n=>dy /dt= [3 (0) *1] - [3 (sin^2 (e^x)) *cos (e^x) ] \n=>dy /dt= -[3 (sin^2 (e^x)) *cos (e^x) ]\n\nI hope that answers your question.\n\nAll the bset
1010282
df(x)/dx = (x)^(2/3)*(2) -2/3(x)^(-1/3)(2x-5) [using product rule]\ndf(x)/dx = 2(x)^(2/3) - 4/3 (x)^(2/3) - 10/3(x)^(-1/3) [simplifying]\ndf(x)/dx = (2/3)(x)^(2/3) - (10/3)(x)^(-1/3)   [ after more simplification]
176854
You can graph it but probably that is not the answer your teacher is looking for. If you wanna do this by the good old method:\n\nf'(x)=3^x*ln(3)\n\nYou are looking for a point where the slope equals to 1 meaning 1 equals to your derivative.\n\n3^x*ln(3)=1\n\nNow use the calculator: x=-0.0856
787521
Example:\nIf z=f(u,v),\nu=g(x,y) and\nv=h(x,y), then\n(all derivatives are supposed to be partial derivatives)\ndz/dx=dz/du*du/dx+dz/dv*dv/dx\ndz/dy=dz/du*du/dy+dz/dv*dv/dy\n\nAlternatively, you can set of the matrices\n[[dz/du dz/dv]]\nand\n[[ du/dx du/dy]\n[dv/dx dv/dy]]\nand get that\n[[dz/dx dz/dy]]\nis the product of the two matrices.