2.7  HIGHER  DERIVATIVES

DEFINITION

The second derivative of a real function f is the derivative of the derivative of f, and is denoted by f ''. The third derivative of f is the derivative of the second derivative, and is denoted by f '', or f(3). In general, the nth derivative of f is denoted by f(n).

If y depends on x, y =f(x), then the second differential of y is defined to be

  d²y = f ′′(x) dx².

In general the nth differential of y is defined by

dny = f(n) (x)dx n.

Here dx² means (dx)² and dxn means (dx) n.

We thus have the alternative notations

For the second and nth derivatives. The notation

y'' = f ''(x),    y(n) = f (n)(x),

is also used.

The definition of the second differential can be remembered in the following way. By definition,

dy= f '(x) dx.

Now hold dx constant and formally apply the Constant Rule for differentiation, obtaining

d(dy) = f ''(x)dx dx,

or                        d²y= f ''(x)dx².

(This is not a correct use of the Constant Rule because the rule applies to a real constant c, and dx is not a real number. It is only a mnemonic device to remember the definition of d²y, not a proof.)

The third and higher differentials can be motivated in the same way. If we hold dx constant and formally use the Constant Rule again and again, we obtain

and so on.

The acceleration of a moving particle is defined to be the derivative of the velocity with respect to time,

a = dv/dt.

Thus the velocity is the first derivative of the distance and the acceleration is the second derivative of the distance. If s is distance, we have

EXAMPLE  1  A ball thrown up with initial velocity b moves according to the equation

y = bt - 16t ²

with y in feet, t in seconds. Then the velocity is

v = b -32 t ft/sec,

and the acceleration (due to gravity) is a constant,

a= -32 ft/sec².

EXAMPLE  2  Find the second derivative of y=sin(2θ).

     First derivative  Put u=2θ. Then

By the Chain Rule,

Second derivative  Let v=2cos (2θ). We must find dv/dθ. Put u= 2 θ. Then

v=2 cos u,

Using the Chain Rule again,

The simplifies to

EXAMPLE  3  A particle moves so that at time t it has gone a distance s along a straight line, its velocity is v, and its acceleration is a. Show that

By definition we have 

So by the Chain Rule,

EXAMPLE  4  If a polynomial of degree n is repeatedly differentiated, the kth derivative will be a    

          polynomial of degree n -k for k ≤ n, and the (n+1)st derivative will be zero. For example,

Geometrically, the second derivative f ′′(x) is the slope of the curve y′= f ′ (x) and is also the rate of change of the slope of the curve y=f(x).

PROBLEMS  FOR  SECTION 2.7

1    y=1/x                             2   y=x5                   3   y=__________

4    f(x)=3x -2                        5   f(x)= x1/2 + x -1/2         6   f(t)= t 3 - 4t 2

7    f(t) =_____                         8   y=(3t - 1)10             9   y=sin x

10   y= cos x                           11   y=A sin (Bx)

12   y= A cos (Bx)                      13   y=e ax

14   y=e -ax                            15  y=1n x

16   y= x 1n x                          17______________

18   y=_________    19_____________    20______________

21   z=________     22 _____________    23_______________

24   Find the third derivative of y= x² - 2/x.

25   A particle moves according to the equation s = 1-1/t², t >0. Find its acceleration.

26   An object moves in such a way that when it has moved a distance s its velocity is v=____ .

     Find its acceleration. (Use Example 3.)

27   Suppose u depends on x and d²u/dx² exists. If y = 3u, find d²y/dx²

28   If d²u/dx² and d² v/dx² exist and y=u+v, find d²x/dx².

29   if d²u/dx² exists and y=u², find d²y/dx².

30   if d²u/dx² and d²v/dx² exist and y=uv, find d²y/dx².

31   Let y=ax2 + bx +c be a polynomial of degree two. Show that dy/dx is a linear function and d²y/dx² is

     a constant function.

□32 Prove that the nth derivative of a polynomial of degree n is constant. (Use the fact that the derivative

     of a polynomial of degree k is a polynomial of degree k-1.)