2x
I'm assuming your question reads "What is the derivative of 3cos(x2)?" You must use the Chain Rule. The derivative of cos(x2) equals -sin(x2) times the derivative of the inside (x2), which is 2x. So... d/dx[3cos(x2)] = -6xsin(x2)
2x is the first derivative of x2.
2x is the first derivative of x2.
-1/x2
-12sin(x3)x2
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
-4/x2
-4 / x3
(-2 x2)' = -4 x
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
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.
The derivative, with respect to x, is -x/sqrt(1-x2)