Product & quotient rules & higher-order

derivatives (2.3) October 17th, 2012

I. the product rule

Thm. 2.7: The Product Rule: The product of two differentiable functions f and g is differentiable. The derivative of fg is the first function times the derivative of the second, plus the second function times the derivative of the first.

d

dx[ f (x)g(x)]= f(x)g'(x)+ g(x) f '(x)

Ex.1: Find the derivative of each function.

(a)h(x)=(5x2 −3x)(4 + 6x)

(b)y=4x3 cosx

(c)y=4xsinx−4cosx

You Try: Find the derivative of each function.

(a) f (x)=(9x−2)(4x2 +1)

(b)g(s)= s(4 −s2 )

(c)g(x)= xsinx

II. the quotient rule

Thm. 2.8: The Quotient Rule: The quotient f/g of two differentiable functions f and g is differentiable for all value of x for which .

g(x)≠0

d

dx

f (x)

g(x)

⎡

⎣⎢

⎤

⎦⎥=

g(x) f '(x)− f(x)g'(x)[g(x)]2

,g(x) ≠0

A. using the quotient rule

Ex. 2: Find the derivative of .y=4x+52x2 −1

You Try: Find the derivative of .f (x)=3x2 −x2x+5

B. Rewriting before differentiating

Ex. 3:Find the slope of the tangent line to the graph of at (-1, -7/3).

f (x)=4 −3 / xx−2

You Try: Find an equation of the tangent line to the graph of at (1, 3).

y=1+2 / x4x−3

*If it is unnecessary to differentiate a function by the quotient rule, it is better to use the constant multiple rule.

c. using the constant multiple rule

Ex. 4: Find the derivative of each function.

(a)

(b)

y=3x2 −x

2

f (x)=2x3 −4x

3x

You Try: Find the derivative of each function.

(a)

(b)

y=4

3x3

g(x)=−2(4x2 −x)

3x

III. derivatives of trigonometric functions

Thm. 2.9: Derivatives of Trigonometric Functions:

d

dx[tan x]=sec2 x

d

dx[cot x]=−csc2 x

d

dx[sec x]=secxtanx

d

dx[csc x]=−cscxcotx

A. Proof of the derivative of sec x

Ex. 5: Prove .d

dx[sec x]=secxtanx

B. differentiating trigonometric functions

Ex. 6: Find the derivative of each function.

(a)

(b)

y=−2x3 −cotx

y=xcscx

C. Different forms of a derivative

Ex. 7:Differentiate both forms of .

y=1+sinxcosx

=secx+ tanx

IV. Higher-order derivatives*We know that we differentiate the position function of an object to obtain the velocity function. We also differentiate the velocity function to obtain the acceleration function. Or, you could differentiate the position function twice to obtain the acceleration function.

s(t) position function

v(t) = s’(t) velocity function

a(t) = v’(t) = s’’(t) acceleration function

*Higher-order derivatives are denoted as follows:

First derivative

Second derivative

Third derivative

Fourth derivative . . .nth derivative

y ', f '(x),dy

dx,d

dx[ f (x)],Dx[y]

y '', f ''(x),d 2y

dx2,d 2

dx2[ f (x)],Dx

2[y]

y ''', f '''(x),d 3y

dx3,d 3

dx3[ f (x)],Dx

3[y]

y(4 ), f (4 )(x),d 4y

dx4,d 4

dx4[ f (x)],Dx

4[y]

y(n), f (n)(x),d ny

dxn,d n

dxn[ f (x)],Dx

n[y]

A. finding acceleration

Ex. 8: Given the position function , find the acceleration at 5 seconds. (Let s(t) be in feet).

s(t)=t+ 32t−2

You Try: Given the position function , where s(t) is in

feet, find that acceleration at 10 seconds.

s(t)=t2 +2t−1

t