-4 / x3
Wiki User
∙ 11y ago2 - 8x
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
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
2x
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).
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
2x is the first derivative of x2.