Calculus: The Language of Change

Exercise Set 6.3
The Product Rule

by Sharon Portz

Exercise 1 : Drill on Products

Break each of the following expressions into a product of two terms and apply the Product Rule to find (dy/dx) for each of the following functions y = y[x].  (The letters a, b, c, and h denote constants or parameters, and e is the natural base for logs and exponentials.

a) y = (2 (x^3) - 4)(3 x + 5)				b) y = ((x^3) - 2 (x^(2 ))+ 4)(5 x + 6)

c) y = ((2/(x^3)) - 4)(3 (Sqrt[x]) + 5)				d) y = ((x^3) - 2 (x^(2 ))+ 3 x + 4)(Sqrt[x])

e) y = ((3/(x^2)) - (4/(Sqrt[x])))(5 (Sqrt[x]) + (3/(x^5)))			f) y = (x + (1/(Sqrt[x])))((Sqrt[x]) + (1/x))

g) y = \(TraditionalForm\`Log[x]\/\@x\%5\)						h) y = ((e^(x )) - 3)((1/(x^4)))

i) y = (x^2)(e^x)						j) y = 7 (e^x) Cos[x]

k) y = (a + b x)(c (x^2) + h (x^3))				l) y = 3 Sin[x]×Cos[7e]

m) y = ((x^((1/2))) + (x^((1/3))) + (x^((1/4))))((x^(-2))+ (x^(-3)) + (x^(-4)))		n) y = ((x^2)+ (x^3) + (x^4))((x^(-(1/2))) + (x^(-(1/3))) + (x^(-(1/4))))

part a

a) y = (2 (x^3) - 4)(3 x + 5)

Solution of part a

part b

b) y = ((x^3) - 2 (x^(2 ))+ 4)(5 x + 6)

Solution of part b

part c

c) y = ((2/(x^3)) - 4)(3 (Sqrt[x]) + 5)

Solution of partc

part d

d) y = ((x^3) - 2 (x^(2 ))+ 3 x + 4)(Sqrt[x])

Solution of partd

part e

e) y = ((3/(x^2)) - (4/(Sqrt[x])))(5 (Sqrt[x]) + (3/(x^5)))

Solution of part e

part f

f) y = (x + (1/(Sqrt[x])))((Sqrt[x]) + (1/x))

Solution of part f

part g

g) y = \(TraditionalForm\`Log[x]\/\@x\%5\)

Solution of part g

part h

h) y = ((e^(x )) - 3)((1/(x^4)))

Solution of part h

\(\(
WalkD[\(\(-3\) + E\^x\)\/x\^4, x]\)\)

part i

i) y = (x^2)(e^x)

Solution of part i

\(\(
WalkD[E\^x\ x\^2, x]\)\)

part j

j) y = 7 (e^x) Cos[x]

Solution of part j

\(\(
WalkD[7\ E\^x\ Cos[x], x]\)\)

part k

k) y = (a + b x)(c (x^2) + h (x^3))

Solution of part k

part l

l) y = 3 Sin[x]×Cos[7e]

Solution of part l

\(\(
WalkD[3\ Cos[7\ E]\ Sin[x], x]\)\)

part m

m) y = ((x^((1/2))) + (x^((1/3))) + (x^((1/4))))((x^(-2))+ (x^(-3)) + (x^(-4)))

Solution of part m

part n

n) y = ((x^2)+ (x^3) + (x^4))((x^(-(1/2))) + (x^(-(1/3))) + (x^(-(1/4))))

Solution of part n

Exercise 2 : Use Product Rule

Which parts of Problem 6.2 can you do now using the Product Rule that you could not do without it? (Again, you can check your work with the DiffRules program by entering the rules up to the Product Rule but not entering the Chain Rule.)

Solution  to exercise 2

The parts of Problem 6.2 where you need the Product Rule are i and n.  Using WalkD the following solutions can be found.

i) y = (Sqrt[x]) Sin[x]

n) y = (e^x) Sin[x]

\(\(
WalkD[E\^x\ Sin[x], x]\)\)

Exercise 3 : h[x] = f[x]g[x]

Let f[x] and g[x] be unknown functions that satisfy f[1] = 2, (df/dx)[1] = 3, g[1] = -3, (dg/dx)[1] = 4.  Let h[x] = f[x]*g[x].  Compute (dh/dx)[1].

Solution

Since h[x] = f[x]*g[x] we can find (dh/dx) using the Product Rule.

(dh/dx) = f[x] d[g[x]] + g[x] d[f[x]]

(dh/dx)[1] = f[1] d[g[1]] + g[1] d[f[1]]

using substitution from what was given we get:

(dh/dx)[1] = 2 * 4 + -3 * 3 = -1

Exercise 4 : Sin[θ] × Cos[θ]

a) Show that the derivative of Sin[θ] × Cos[θ]  is Cos[2θ] using the addition formula for cosine.  (You can look that formula up in Chapter 28 on the CD.)

b) Combine this fact with the previous example to show that

((d(Sin[2 x]))/dx) = ((d(2 Sin[x] Cos[x]))/dx) = 2 Cos[2 x]

without using the Chain Rule! (We will be able to verify this more simply once we have the Chain Rule.  The point is that there are often several ways to apply the rules of algebra and trig in combination with the rules of differentiation.)

part a

a) Show that the derivative of Sin[θ] × Cos[θ]  is Cos[2θ] using the addition formula for cosine.  (You can look that formula up in Chapter 28 on the CD.)

Solution  to part a

The formula we need from Chapter 28 (28.9 to be exact) is

Cos[x + y] = Cos[x]Cos[y] - Sin[x]Sin[y]

Cos[2θ] = Cos[θ + θ] = Cos[θ]Cos[θ] - Sin[θ]Sin[θ] = (((Cos[θ]))^2) - (((Sin[θ]))^2)

So we have to show that the derivative of Sin[θ] × Cos[θ] =(((Cos[θ]))^2) - (((Sin[θ]))^2)

Using WalkD we see:

Since Cos[θ]^2 - Sin[θ]^2 is another way to write (((Cos[θ]))^2) - (((Sin[θ]))^2), the derivative ofSin[x]Cos[x] is Cos[2θ].

part b

b) Combine this fact with the previous example to show that

((d(Sin[2 x]))/dx) = ((d(2 Sin[x] Cos[x]))/dx) = 2 Cos[2 x]

without using the Chain Rule! (We will be able to verify this more simply once we have the Chain Rule.  The point is that there are often several ways to apply the rules of algebra and trig in combination with the rules of differentiation.)

Solution to part b

In section 9 of chapter 28 we get the following formula:

Sin[x + y] = Sin[x]Cos[y] + Sin[y]Cos[x]

Therefore, Sin[2x] = 2 * Sin[x]Cos[x]

Thus, ((d(Sin[2 x]))/dx)= ((d(2 Sin[x] Cos[x]))/dx) and using WalkD again we get:

From above we know (Cos[x]^2 - Sin[x]^2) = Cos[2x]  therefore

((d(2 Sin[x] Cos[x]))/dx) = 2 Cos[2 x].

Exercise 5 : Other Variables

a) v = (e^u)Cos[u]
	b) v = ((u^3)+ 3u + 3)(Sin[u] + Cos[u])

c) u = (4 (Sqrt[v]) - π (1/(v^2)))((v^((1/4))) - (1/(v^((1/3)))))
	d) y = (Cos[v] - Sin[v])(Log[v] - (e^v))

part a

a) v = (e^u)Cos[u]

Solution of part a

Using WalkD:

\(\(
WalkD[E\^u\ Cos[u], u]\)\)

part b

b) v = ((u^3)+ 3u + 3)(Sin[u] + Cos[u])

Solution of part b

part c

c) u = (4 (Sqrt[v]) - π (1/(v^2)))((v^((1/4))) - (1/(v^((1/3)))))

Solution of part c

part d

d) y = (Cos[v] - Sin[v])(Log[v] - (e^v))

Solution of part d

\(\(
WalkD[\((\(-E\^v\) + Log[v])\)\ \((Cos[v] - Sin[v])\), v]\)\)