Second Derivative?

Question

What does the second derivative of a function represent relative to the first derivative? Explain.

This is the last question on my homework and it has me flummoxed!
If Im correct the first derivative represents the rate of change of y relative to x. So does that mean the second derivative represents the rate of change of the rate of change (aka the acceleration of change?) Please use F(x)=x^2 and help me out! :'(

Answer 1

Yes, your thinking is correct.

y = x²
dy/dx = 2x
d²y/dx² = 2

For an equation dealing with motion, the first derivative will give you the speed of the object, that is, the rate of distance with respect to time.
The second derivative of the equation will give you the acceleration of the object, that is, the rate of change of speed with respect to time.

Derivatives can also be used for maximum and minimum calculations. Example, to calculate the maximum volume of a can using the minimum amount of steel.

It's a powerful piece of maths.

Answer 2

You're right. The second derivative is simply the rate of change of the function represented by the first derivative.

Answer 3

you are correct for example the first derivative of a distance against time graph gives the velocity which is the rate of change of distance against time then the second derivative is the rate of change of velocity against time which gives the acceleration.

first derivative of x^2 is 2x

second derivative is the derivative of 2x which is 2

Answer 4

The 2nd derivative represents the rate of change, or slope, of the 1st derivative
given
y = x^2
y' = 2x, the slope of x^2 at any x value
y'' = 2, the slope of 2x

Answer 5

yes
the first derivative represents the rate of change of y relative to x

the second derivative represents the rate of change of the rate of change of y relative to x to x (that mean second)

example
s(t) is distant at time
the first derivative is ds/dt that mean v(t) at time
the second derivative of s(t) mean derivative v(t)
it same dv/dt like a(t) concept
y=x^2
f(x) = x^2
dy/dx = f'(x) = 2x
d^2y/dx^2 = f''(x) = 2

f'(x) = lim h->0 (f(x+h)-f(x))/h
f''(x) = lim h->0 (f'(x+h)-f'(x))/h

Answer 6

If y = f(t) is distance and t is time:-
dy / dt = f `(t) = velocity at time t
d²y / dt² = f "(t) = acceleration at time t
Example
y is distance in metres.
t is time in seconds.
y = f(t) = t² + t
dy / dt = f `(t) = 2t + 1
d²y / dt² = f "(t) = 2
After 10 seconds v = 21 m/sec
Acceleration is constant at 2 m / sec²
Distance = 110 m

Hope this helps.