How do you differentiate: sqrt(x)e^(6x^2)(x^2+1)^10?
827232
d/dx of rtx=1/2rtx\nd/dx of (6x^2)=12x\nd/dx of (x^2+1)^10\n=10(x^2+1)^9*2x\nd/dx of rtx(6x^2)(x^2+1)^10\nd/dx of rtx[(6x^2)(x^2+1)^10]\n+rtx*d/dx(6x^2)(x^2+1)^10\nd/dx of (6x^2)(x^2+1)^10\n=6x^2*d/dx(x^2+1)^10\n+(x^2+1)^10*d/dx of 6x^2\nrest i think you can complete
751318
x^2-x-6>0\nor, x^2-3x+2x-6>0\nor, x(x-3)+2(x-3)>0\nor, (x-3)(x+2)>0\nor x>3 and x<2\n\nhope this helps :)
643228
This is unclear.  Use the symbol ^ when using exponents.\nDo you mean?\n(6x10^12)^12 times (3.2x10^8) divided by [(1.6x10^7)^3 times 10^2]?
766061
p^9-q^9/p^6-q^6\n=(p^3-q^3)(p^6+p^3q^3+q^6)/(p^3+q^3)(p^3-q^3)\n=(p^6+p^3q^3+q^6)/(p^3+q^3)
198537
arctan'(x) = 1/(1+x^2), so differentiating using the good old rules, we get the result (using 'sq' for the square root to avoid yahoo cutting the equation short):\n\nsq(a+x^2)-1\n-------------------------------\n(x*sq(a+x^2)(1+(x-sq(a+x^2))^2\n\nMan, that's ugly. :-)
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
120236
6x\n\n3x^2 --> 3*2 x^2-1 --> 6x\n\nUpdate: You are deriving with respect to x, right?
903727
-1 / ( 20 ( ln( 10 + e ) * ( 10 + e )^(10 * x^2 ) ) )
706332
The antiderivative is:\n\n(1/2)erfi(sqrt(pi)*x)\n\nI used Wolfram
720929
Let me just work through the steps to solve the problem:\n\nsqrt(2x^2-1)=x   <---Original Equation\n2x^2-1=x         <---Square both sides\n2x^2-x-1=0       <---Subtract x\n\nNow, apply the quadradic equation:\nx=(-b+/-(sqrt(b^2-4ac))/(2a)\n\nFor this equation, the equation is:\nx=(-(-1)+/-(sqrt((-1^2)-4*2*(-1))/(2*(2))\n\nSimplified it is:\nx=(1+/-sqrt(1+8))/(4)\nx=(1+3)/4 or x=(1-3)/4\n\nSo, x=1 or x=-1/2
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
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.
97125
use the distance formula\nsqrt(3^2+(-sqr3)^2)=?
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)
914223
difference of squares rule:\n\nx^2 - y^2 \n\n(x-y)(x+y)
606423
Here's how to work it out:\n\n(Sorry, because I can't use special symbols, I'm using _/ for the square root symbol and <= for the less than or equal to symbol.)\n\n_/3x + 6 + 4 <= 7\n_/3x + 10 <= 7\n3x + 10 <= 7 x 7\n3x + 10 <= 49\n3x + 10 - 10 <= 49 - 10\n3x <= 39\n3x/3 <= 39/3\nx <= 13
866864
Just differentiate it. You need f'(x) < 0.\n\nBy the quotient rule, f'(x) = (2x(x^2 + 4) - (x^2)(2x))/(x^2+4)^2.\nNow, since the bottom is a square, thus positive, you just need the numerator to be < 0:\n2x(x^2 + 4) - x^2 * 2x < 0\n2x^3 + 8x - 2x^3 < 0\n8x < 0\nx < 0\nThus it is decreasing on (-∞,0).