f(x) = 83 - x2
Difference Quotient
You are going to plug in f(x) into:
f '(x) lim Δx -> 0 = f(x + Δx) - f(x) / Δx
= 83 - (x + Δx)2 - (83 - x2) / Δx
= 83 - (x2 + 2xΔx + (Δx)2) - 83 + x2 / Δx
= 83 - x2 - 2xΔx - (Δx)2 - 83 + x2 / Δx
The 83 and x2 cancel out leaving:
= - 2xΔx - (Δx)2 / Δx
Through factoring out the Δx:
= Δx (-2x - Δx) / Δx
The Δx cancel out leaving:
= -2x - Δx
The limit of Δx is 0, thus plug in 0 and you get
= -2x - 0
= -2x
Power Rule
The power rule is simply to bring down the exponent and subtract one from the remaining exponent. Changes are done in bold. Reminder: the derivative of a constant is zero.
f'(x) = 0 - (2)x2-1
= -2x
-1/x2
-4/x2
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
0
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
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
2x is the first derivative of x2.
2x is the first derivative of 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
-1/x2
-12sin(x3)x2
(-2 x2)' = -4 x
-4/x2
-4 / x3
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.