Following the correct order of operations: derivative of x^2 + 6/2 = derivative of x^2 +3, which equals 2x
The remainder is 8. (x2 + 4)/(x - 2) = (x + 2) + 8/(x - 2) or x2 + 4 = (x - 2)(x + 2) + 8
(3x4 + 2x3 - x2 - x - 6)/(x2 + 1)= 3x2 + 2x - 4 + (-3x - 2)/(x2 + 1)= 3x2 + 2x - 4 - (3x + 2)/(x2 + 1)where the quotient is 3x2 + 2x - 4 and the remainder is -(3x + 2).
x4 +x2 =x2 (x2+1)
All it means to take the second derivative is to take the derivative of a function twice. For example, say you start with the function y=x2+2x The first derivative would be 2x+2 But when you take the derivative the first derivative you get the second derivative which would be 2
(x2 + 14x + 49) = (x + 7)2
The anti-derivative of X2 plus X is the same as the anti-derivative of X2 plus the anti-derivative of X. The anti derivative of X2 is X3/3 plus an integration constant C1 The anti derivative of X is X2/2 plus an integration constant C2 So the anti-derivative of X2+X is (X3/3)+(X2/2)+C1+C2 The constants can be combined and the fraction can combined by using a common denominator leaving (2X3/6)+(3X2/6)+C X2/6 can be factored out leaving (X2/6)(2X+3)+C Hope that helps
Well if you have 5/X then you can rewrite this like 5x-1. And the derivative to that is -5x-2 and that can be rewrote to: -(5/x2).
The antiderivative of x/(x2-1) is ln(x2-1)/2. Proof: (ln(x2-1)/2)' = (1/(x2-1))*(x2-1)'/2=1/(x2-1)*(2x/2)=x/(x2-1).
d/dx (x2+ 9)1/2= 1/2*(x2+ 9)-1/22x = x(x2+ 9)-1/2or x/(x2+ 9)1/2
x3+3x2+3x+2 divided by x+2 equals x2+x+1
(-2 x2)' = -4 x
The remainder is 8. (x2 + 4)/(x - 2) = (x + 2) + 8/(x - 2) or x2 + 4 = (x - 2)(x + 2) + 8
Answer is x2 -6x+14 with remainder 2
(3x4 + 2x3 - x2 - x - 6)/(x2 + 1)= 3x2 + 2x - 4 + (-3x - 2)/(x2 + 1)= 3x2 + 2x - 4 - (3x + 2)/(x2 + 1)where the quotient is 3x2 + 2x - 4 and the remainder is -(3x + 2).
-1
x4 +x2 =x2 (x2+1)
Any monomial in the format: axn has a derivative equal to: nax(n - 1) In this case, "a" is equal to 1 and "n" is equal to 2. So the derivative of x2 is equal to 2x.