Two curves are said to be orthogonal at each point (x0, y0) of their intersection if their tangent lines are perpendicular. Show that the following families of curves are orthogonal.

11.x2 + y2 = r2, y + mx = 0

12.(x − h)2 + y2 = h2, x2 + (y − k)2 = k2

3.5Higher Order Derivatives

If the vertical height y of an object is a function f of time t, then y0(t) is called its velocity, denoted v(t). The derivative v0(t) is called the acceleration of the object and is denoted a(t). That is,

y(t) = f(t), y0(t) = v(t), v0(t) = a(t).

We say that a(t) is the second derivative of y, with respect to t, and write

Derivatives of order two or more are called higher derivatives and are represented by the following notation:

CHAPTER 3. DIFFERENTIATION

f(n)(x) = dnf dxn

which is read as “the nth derivative of f with respect to x.”

Part(iv) y0 = −5 csc2(5x), y00 = −10 csc(5x)[(− csc 5x cot 5x) · 5]

y00 = 50 csc2(5x) cot(5x)

Part(v) y0 = 2 sec(2x) tan(2x)

y00 = 2[(2 sec(2x) tan(2x)) · tan(2x) + sec(2x) · (2 sec2(2x))]

y00 = 4 sec(2x) tan2(2x) + 4 sec3(2x)

Part(vi) y0 = −2x csc(x2) cot(x2)

y00 = −2[1 · csc(x2) cot(x2) + x(−2x csc(x2) cot(x2)) · cot(x2)

+x csc(x2) · (−2x csc2(x2))]

=−2 csc(x2) cot(x2) + 4x2 csc(x2) cot2(x2) + 4x2 csc3(x2)

Example 3.5.2 Compute the second order derivative of each of the following functions:

Part (i) y0 = 3 cosh(3x), y00 = 9 sinh(3x)

Part (ii) y0 = 2x sinh(x2), y00 = 2 sinh(x2) + 2x(2x cosh x2) or

y00 = 2 sinh(x2) + 4x2 cosh(x2)

Part (iii) y0 = 2 sech2(2x), y00 = 2 · (2 sech(2x) · (−sech(2x) tanh(2x) · 2)),

y00 = −8 sech2(2x) tanh(2x)

Part (iv) y0 = −4 csch2(4x), y00 = −4(2(csch(4x)) · (−csch(4x) coth(4x) · 4))

y00 = 32 csch2(4x) coth(4x)

Part (v) y0 = −5 sech (5x) tanh(5x)

y0 = −5[−5 sech(5x) tanh(5x) · tanh(5x) + sech(5x) · sech2(5x) · 5]

y0 = 25 sech(5x) tanh2(5x) − 25 sech3(5x).

Part (vi) y0 = −10 csch(10x) coth(10x)

y00 = −10[−10 csch(10x) coth(10x) · coth(10x)

Example 3.5.3 Compute the second order derivatives for the following functions:

Part (i) y0 = x2x2 = x2 = 2x−1

y00 = −2x−2 = −x22.

Part (ii) y0 = 2xex2 , y00 = 2ex2 + 4x2ex2 = (2 + 4x2)ex2 .

Part (iv) y0 = 10x2 · (ln 10) · 2x

y00 = 2 ln 10[10x2 + x · 10x2 ln 10 · 2x]

y00 = 10x2 [2 ln 10 + (2 ln 10)2x2]

Example 3.5.4 Compute the second derivatives of the following functions:

Then

Part (i)

= −21 (2x)(1 + x2)−3/2

x

= −(1 + x2)3/2 .

Example 3.5.5 Find y00 for the equation x2 + y2 = 4.

First, we find y0 by implicit di erentiation.

2x + 2yy0 = 0 → y0, xy .

Now, we di erentiate again with respect to x.

Example 3.5.6 Compute y00 for x3 + y3 = 4xy.

From Example 25 in the last section we found that

To find y00, we di erentiate y0 with respect to x to get

y00 = (3y2 − 4x)(4y0 − 3x2) − (4y − 3x2)(6yy0 − 4) , 3y2 − 4x 6= 0. 3y2 − 4x

In order to simplify any further, we must first replace y0 by its computed value. We leave this as an exercise.

Example 3.5.7 Compute f(n)(c) for the given f and c and all natural numbers n:

To compute the general nth derivative formula we must discover a pattern and then generalize the pattern.

Part (i) f(x) = sin x, f0(x) = cos x, f00(x) = − sin x, f000(x) = cos x, f4(x) = sin x. Then the next four derivatives are repeated and so on. We get

f(4n)(n) = sin x, f(4n+1)(x) = cos x, f(4n+2)(x) = − sin x, f(4n+3)(x) = − cos x.

By evaluating these at c = 0, we get

f(4n)(0) = 0, f(4n+2)(0); f(4n+1)(0) = 1 and f(4n+3)(0) = −1,

for n = 0, 1, 2, · · ·

Part (ii) This part is similar to Part (i) and is left as an exercise.

Part (iii) f(x) = ln x, f0(x) = x−1, f00(x) = (−1)x−2, f(3)(x) = (−1)(−2)x−3, . . . ., f(n)(x) = (−1)(−2) . . . (−(n − 1))x−n = (−1)n−1(n − 1)!x−n, f(n)(1) = (−1)n−1(n − 1)!, n = 1, 2, . . .