y = 3 + 2x - 4x^(2) dy/dx = 2 - 8x ( The derivative).
1/(1-x2 )1/2
-1/(2*x2)
d/dx(lnx2) = 2/x ===== d/dx2(2/x) use product rule 2'*x - 2 *x'/x2 = - 2/x2 -----------
d/dx (x2+ 9)1/2= 1/2*(x2+ 9)-1/22x = x(x2+ 9)-1/2or x/(x2+ 9)1/2
X/1 is just X. so (1/2)X2 + C or X2/2 + C
Derivative of 1/x 1/x = x-1 Take the derivative (-1)x(-1-1) = -x-2 = 1/x2
(-2 x2)' = -4 x
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
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)
Following the correct order of operations: derivative of x^2 + 6/2 = derivative of x^2 +3, which equals 2x
1/x = x-1d/dx(x-1) = -x-2 = -1/x2
2
-1
y = 3 + 2x - 4x^(2) dy/dx = 2 - 8x ( The derivative).
1/(1-x2 )1/2
2x