To differentiate logarithmic functions with bases other than e, use

1 To differentiate logarithmic functions with bases other than e, use

log_b m = ln m lnb

log_b m = ln m lnb Example Find the derivative of y = log₂(5x3).

1 To differentiate logarithmic functions with bases other than e, use

log_b m = ln m lnb Example Find the derivative of y = log₂(5x3).

Solution dy

dx =

d dx

ln 5x3 ln 2

log_b m = ln m lnb Example Find the derivative of y = log₂(5x3).

Solution dy

dx =

d dx

ln 5x3 ln 2

!

= d dx

1

ln 2 · ln 5x 3

1 To differentiate logarithmic functions with bases other than e, use

log_b m = ln m lnb Example Find the derivative of y = log₂(5x3).

Solution dy

dx =

d dx

ln 5x3 ln 2

!

= d dx

1

ln 2 · ln 5x 3

!

= 1 ln 2 ·

d

log_b m = ln m lnb Example Find the derivative of y = log₂(5x3).

Solution dy

dx =

d dx

ln 5x3 ln 2

!

= d dx

1

ln 2 · ln 5x 3

!

= 1 ln 2 ·

d

dx(ln 5 + 3 ln x) properties of log = 1

ln 2 0 + 3 · 1 x

1 To differentiate logarithmic functions with bases other than e, use

log_b m = ln m lnb Example Find the derivative of y = log₂(5x3).

Solution dy

dx =

d dx

ln 5x3 ln 2

!

= d dx

1

ln 2 · ln 5x 3

!

= 1 ln 2 ·

d

dx(ln 5 + 3 ln x) properties of log = 1

ln 2 0 + 3 · 1 x

!

= 3 xln 2

Exponential functions of other bases To differentiate f(x) = bx where b , e

2

Exponential functions of other bases To differentiate f(x) = bx where b , e

Method 1 Express bx using exponential with base e.

Exponential functions of other bases To differentiate f(x) = bx where b , e

Method 1 Express bx using exponential with base e. y = bx

2

Exponential functions of other bases To differentiate f(x) = bx where b , e

Method 1 Express bx using exponential with base e. y = bx

ln y = lnbx = xlnb

Exponential functions of other bases To differentiate f(x) = bx where b , e

Method 1 Express bx using exponential with base e. y = bx

ln y = lnbx = xlnb y = e(lnb)x

2

Exponential functions of other bases To differentiate f(x) = bx where b , e

Method 1 Express bx using exponential with base e. y = bx

ln y = lnbx = xlnb y = e(lnb)x

Method 2 Use a technique called logarithmic differentiation. Both methods need chain rule.

Chapter 9: More Differentiation

• Implicit Differentiation

• More Curve Sketching

• More Extremum Problems

Objectives

• To use Chain Rule to do differentiation.

• To use Implicit Differentiation to find dy

dx.

4

Up to this moment, can differentiate “simple” functions like (1) f(x) = x5 + 1

(2) f(x) = x − 1 x + 1 (3) f(x) = sin x

(4) f(x) = ex + 2 tan x (5) f(x) = ln x

cos x −

ex x2 _{+ 1}

How about

(1) g(x) = sin(x2) ?

(2) g(x) = ex2+1 ?

5

How about

(1) g(x) = sin(x2) ?

(2) g(x) = ex2+1 ?

(3) g(x) = ln(1 + 2x) ?

Need the chain rule – most important rule for finding derivatives, used for differentiating composite functions.

How about

(1) g(x) = sin(x2) ? g(x) = sin f(x) where f(x) = x2

(2) g(x) = ex2+1 ?

(3) g(x) = ln(1 + 2x) ?

Need the chain rule – most important rule for finding derivatives, used for differentiating composite functions.

5

How about

(1) g(x) = sin(x2) ? g(x) = sin f(x) where f(x) = x2

(2) g(x) = ex2+1 ? g(x) = ef(x) where f(x) = x2 + 1

(3) g(x) = ln(1 + 2x) ?

Need the chain rule – most important rule for finding derivatives, used for differentiating composite functions.

How about

(1) g(x) = sin(x2) ? g(x) = sin f(x) where f(x) = x2

(2) g(x) = ex2+1 ? g(x) = ef(x) where f(x) = x2 + 1

(3) g(x) = ln(1 + 2x) ? g(x) = ln f(x) where f(x) = 1 + 2x

Need the chain rule – most important rule for finding derivatives, used for differentiating composite functions.

6 Composition of Functions

Recall (g ◦ f)(x) = gf(x)

Example Let f(x) = sin x and g(x) = x2. Find (1) (f ◦ g)(x)

6 Composition of Functions

Recall (g ◦ f)(x) = gf(x)

Example Let f(x) = sin x and g(x) = x2. Find (1) (f ◦ g)(x)

Solution (f ◦ g)(x) = fg(x)

Recall (g ◦ f)(x) = gf(x)

Example Let f(x) = sin x and g(x) = x2. Find (1) (f ◦ g)(x)

Solution (f ◦ g)(x) = fg(x) = f(x2)

6 Composition of Functions

Recall (g ◦ f)(x) = gf(x)

Example Let f(x) = sin x and g(x) = x2. Find (1) (f ◦ g)(x)

Solution (f ◦ g)(x) = fg(x) = f(x2)

= sin(x2) = sin x2 (2) (g ◦ f)(x)

Recall (g ◦ f)(x) = gf(x)

Example Let f(x) = sin x and g(x) = x2. Find (1) (f ◦ g)(x)

Solution (f ◦ g)(x) = fg(x) = f(x2)

= sin(x2) = sin x2 (2) (g ◦ f)(x)

6 Composition of Functions

Recall (g ◦ f)(x) = gf(x)

Example Let f(x) = sin x and g(x) = x2. Find (1) (f ◦ g)(x)

Solution (f ◦ g)(x) = fg(x) = f(x2)

= sin(x2) = sin x2 (2) (g ◦ f)(x)

Solution (g ◦ f)(x) = gf(x) = g(sin x)

Recall (g ◦ f)(x) = gf(x)

Example Let f(x) = sin x and g(x) = x2. Find (1) (f ◦ g)(x)

Solution (f ◦ g)(x) = fg(x) = f(x2)

= sin(x2) = sin x2 (2) (g ◦ f)(x)

Solution (g ◦ f)(x) = gf(x) = g(sin x)

7

Composition of Functions Recall (g ◦ f)(x) = gf(x)

Composition of Functions Recall (g ◦ f)(x) = gf(x)

Example Express ex2+1 as composition of two functions. Solution Let f(x) = x2 + 1 and

7

Composition of Functions Recall (g ◦ f)(x) = gf(x)

Example Express ex2+1 as composition of two functions. Solution Let f(x) = x2 + 1 and

Composition of Functions Recall (g ◦ f)(x) = gf(x)

Example Express ex2+1 as composition of two functions. Solution Let f(x) = x2 + 1 and g(x) = ex. Then

7

Composition of Functions Recall (g ◦ f)(x) = gf(x)

Example Express ex2+1 as composition of two functions. Solution Let f(x) = x2 + 1 and g(x) = ex. Then

ex2+1 = ef(x) = gf(x)

8

Chain Rule If y is a differentiable function of u and u is a differentiable function of x, then y is a differentiable function of x and

dy dx =

dy du ·

du dx

of x, then y is a differentiable function of x and dy

dx = dy du ·

du dx Idea of proof dy

dx = ∆xlim→0 ∆y

∆x limh→0

f(x + h) − f(x)

8

Chain Rule If y is a differentiable function of u and u is a differentiable function of x, then y is a differentiable function of x and

dy dx =

dy du ·

du dx Idea of proof dy

dx = ∆xlim→0 ∆y

∆x limh→0

f(x + h) − f(x)

h

= lim

∆x→0

∆y

∆u ·

∆u

∆x

of x, then y is a differentiable function of x and dy

dx = dy du ·

du dx Idea of proof dy

dx = ∆xlim→0 ∆y

∆x limh→0

f(x + h) − f(x)

h

= lim

∆x→0

∆y

∆u ·

∆u

∆x

!

= lim

∆u→0

∆y

∆u · ∆x→0lim ∆u

8

Chain Rule If y is a differentiable function of u and u is a differentiable function of x, then y is a differentiable function of x and

dy dx =

dy du ·

du dx Idea of proof dy

dx = ∆xlim→0 ∆y

∆x limh→0

f(x + h) − f(x)

h

= lim

∆x→0

∆y

∆u ·

∆u

∆x

!

= lim

∆u→0

∆y

∆u · ∆x→0lim ∆u ∆x = dy du · du dx

Chain Rule in alternative form Put y = f(u) and u = g(x).

9

Chain Rule in alternative form Put y = f(u) and u = g(x).

Chain Rule in alternative form Put y = f(u) and u = g(x).

Then y = (f ◦ g)(x) (composition of functions) (f ◦ g)0(x) = dy

9

Chain Rule in alternative form Put y = f(u) and u = g(x).

Then y = (f ◦ g)(x) (composition of functions) (f ◦ g)0(x) = dy

dx

= dy

du · du dx

Chain Rule in alternative form Put y = f(u) and u = g(x).

Then y = (f ◦ g)(x) (composition of functions) (f ◦ g)0(x) = dy

dx

= dy

du · du dx

9

Chain Rule in alternative form Put y = f(u) and u = g(x).

Then y = (f ◦ g)(x) (composition of functions) (f ◦ g)0(x) = dy

dx

= dy

du · du dx

= f0(u) · g0(x) (f ◦ g)0(x) = f0 g(x) · g0(x)

dx

(1) without using chain rule; (2) using chain rule.

10 Example Find d

dx(x

2 _{+ 5)}3

(1) without using chain rule; (2) using chain rule.

Solution

(1) (without chain rule) Expanding

(x2 + 5)3 = (x2)3 + 3(x2)2(5) + 3(x2)(52) + 53 = x6 + 15x4 + 75x2 + 125

dx

(1) without using chain rule; (2) using chain rule.

Solution

(1) (without chain rule) Expanding

(x2 + 5)3 = (x2)3 + 3(x2)2(5) + 3(x2)(52) + 53 = x6 + 15x4 + 75x2 + 125

Differentiating term by term: d

dx(x

2 _{+ 5)}3 ₌ d

dx(x

6 _{+ 15x}4 _{+ 75x}2 _{+ 125)}

= 6x5 + 15 · 4x3 + 75 · 2x = 6x5 + 60x3 + 150x

11 Example Find d

dx(x

2 _{+ 5)}3

(2) using chain rule.

Solution

dx

(2) using chain rule.

Solution

(2) (using chain rule)

11 Example Find d

dx(x

2 _{+ 5)}3

(2) using chain rule.

Solution

(2) (using chain rule)

dx

(2) using chain rule.

Solution

(2) (using chain rule)

11 Example Find d

dx(x

2 _{+ 5)}3

(2) using chain rule.

Solution

(2) (using chain rule)

• Put u = x2 + 5 and y = u3. Then y = (x2 + 5)3.

• d

dx(x

2 _{+ 5)}3 ₌ dy

dx

(2) using chain rule.

Solution

(2) (using chain rule)

• Put u = x2 + 5 and y = u3. Then y = (x2 + 5)3.

• d

dx(x

2 _{+ 5)}3 ₌ dy

dx

= dy

du · du

11 Example Find d

dx(x

2 _{+ 5)}3

(2) using chain rule.

Solution

(2) (using chain rule)

• Put u = x2 + 5 and y = u3. Then y = (x2 + 5)3.

• d

dx(x

2 _{+ 5)}3 ₌ dy

dx

= dy

du · du

dx chain rule

= d

duu

3 _· d dx(x

dx

(2) using chain rule.

Solution

(2) (using chain rule)

• Put u = x2 + 5 and y = u3. Then y = (x2 + 5)3.

• d

dx(x

2 _{+ 5)}3 ₌ dy

dx

= dy

du · du

dx chain rule

= d

duu

3 _· d dx(x

2 _{+ 5)} = 3u2 · 2x

11 Example Find d

dx(x

2 _{+ 5)}3

(2) using chain rule.

Solution

(2) (using chain rule)

• Put u = x2 + 5 and y = u3. Then y = (x2 + 5)3.

• d

dx(x

2 _{+ 5)}3 ₌ dy

dx

= dy

du · du

dx chain rule

= d

duu

3 _· d dx(x

2 _{+ 5)} = 3u2 · 2x

dx

(2) using chain rule.

Solution

(2) (using chain rule)

• Put u = x2 + 5 and y = u3. Then y = (x2 + 5)3.

• d

dx(x

2 _{+ 5)}3 ₌ dy

dx

= dy

du · du

dx chain rule

= d

duu

3 _· d dx(x

2 _{+ 5)} = 3u2 · 2x

= 3(x2 + 5)2(2x) = 6x(x2 + 5)2

12

Method 1 Answer is 6x5 + 60x3 + 150x

Method 2 Answer is 6x(x2 + 5)2 Remark 1

Remark 1

• The above two results are the same.

Remark 2

12

Method 1 Answer is 6x5 + 60x3 + 150x

Method 2 Answer is 6x(x2 + 5)2 Remark 1

• The above two results are the same.

Remark 2

• If change the function to y = (x2 + 5)13, first method can’t be applied.

Remark 1

• The above two results are the same.

Remark 2

• If change the function to y = (x2 + 5)13, first method can’t be applied.

Method 1 (x2 + 5)3 = (x2)3 + 3(x2)2(5) + 3(x2)(52) + 53 (x2 + 5)13 no way to expand

12

Method 1 Answer is 6x5 + 60x3 + 150x

Method 2 Answer is 6x(x2 + 5)2 Remark 1

• The above two results are the same.

Remark 2

• If change the function to y = (x2 + 5)13, first method can’t be applied.

Method 1 (x2 + 5)3 = (x2)3 + 3(x2)2(5) + 3(x2)(52) + 53 (x2 + 5)13 no way to expand

Remark 1

• The above two results are the same.

Remark 2

• If change the function to y = (x2 + 5)13, first method can’t be applied.

Method 1 (x2 + 5)3 = (x2)3 + 3(x2)2(5) + 3(x2)(52) + 53 (x2 + 5)13 no way to expand

Method 2 Put u = x2 + 5 and y = u3. Then y = (x2 + 5)3. Put u = x2 + 5 and y = u31. Then y = (x2 + 5)13.

12

Method 1 Answer is 6x5 + 60x3 + 150x

Method 2 Answer is 6x(x2 + 5)2 Remark 1

• The above two results are the same.

Remark 2

• If change the function to y = (x2 + 5)13, first method can’t be applied.

Method 1 (x2 + 5)3 = (x2)3 + 3(x2)2(5) + 3(x2)(52) + 53 (x2 + 5)13 no way to expand

Method 2 Put u = x2 + 5 and y = u3. Then y = (x2 + 5)3. Put u = x2 + 5 and y = u31. Then y = (x2 + 5)13. • Second method makes use of

the chain rule

Simple Form General Form d

dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x) d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] · d

dx f(x) d

dx cos x = −sin x

d

dx cos[f(x)] = −sin[f(x)] · d

dx f(x) d

dx tan x = sec

2 _x d

dx tan[f(x)] = sec

2_{[f(x)] ·} d

dx f(x) d

dxe

x _{= e}x d

dxe

f(x) _{= e}f(x) _· d

dx f(x) d

dx ln x = 1 x

d

dx ln[f(x)] = 1 f(x) ·

d

14

(3) d

dx cos[f(x)] = −sin[f(x)] · d

dx f(x) Proof

(3)

dx cos[f(x)] = −sin[f(x)] · dx f(x) Proof Put u = f(x) and y = cosu.

14

(3) d

dx cos[f(x)] = −sin[f(x)] · d

dx f(x)

(3)

dx cos[f(x)] = −sin[f(x)] · dx f(x)

Proof Put u = f(x) and y = cosu. Then y = cos f(x) and so

d

dxcos[f(x)] =

dy

14

(3) d

dx cos[f(x)] = −sin[f(x)] · d

dx f(x)

Proof Put u = f(x) and y = cosu. Then y = cos f(x) and so

d

dxcos[f(x)] =

dy

dx

= dy

du · du

(3)

dx cos[f(x)] = −sin[f(x)] · dx f(x)

Proof Put u = f(x) and y = cosu. Then y = cos f(x) and so

d

dxcos[f(x)] =

dy

dx

= dy

du · du

dx chain rule

= d

du cosu · du dx

14

(3) d

dx cos[f(x)] = −sin[f(x)] · d

dx f(x)

Proof Put u = f(x) and y = cosu. Then y = cos f(x) and so

d

dxcos[f(x)] =

dy

dx

= dy

du · du

dx chain rule

= d

du cosu · du dx

= −sinu · du

(3)

dx cos[f(x)] = −sin[f(x)] · dx f(x)

Proof Put u = f(x) and y = cosu. Then y = cos f(x) and so

d

dxcos[f(x)] =

dy

dx

= dy

du · du

dx chain rule

= d

du cosu · du dx

= −sinu · du

dx

= −sin[f(x)] · d

15

(6) d

dx ln[f(x)] = 1 f(x) ·

d

dx f(x) Proof

(6)

dx ln[f(x)] = f(x) · dx f(x) Proof Put u = f(x) and y = lnu.

15

(6) d

dx ln[f(x)] = 1 f(x) ·

d

dx f(x)

(6)

dx ln[f(x)] = f(x) · dx f(x)

Proof Put u = f(x) and y = lnu. Then y = ln f(x) and so

d

dxln[f(x)] =

dy

15

(6) d

dx ln[f(x)] = 1 f(x) ·

d

dx f(x)

Proof Put u = f(x) and y = lnu. Then y = ln f(x) and so

d

dxln[f(x)] =

dy

dx

= dy

du · du

(6)

dx ln[f(x)] = f(x) · dx f(x)

Proof Put u = f(x) and y = lnu. Then y = ln f(x) and so

d

dxln[f(x)] =

dy

dx

= dy

du · du

dx chain rule

= d

du lnu · du dx

15

(6) d

dx ln[f(x)] = 1 f(x) ·

d

dx f(x)

Proof Put u = f(x) and y = lnu. Then y = ln f(x) and so

d

dxln[f(x)] =

dy

dx

= dy

du · du

dx chain rule

= d

du lnu · du dx = 1 u · du dx

(6)

dx ln[f(x)] = f(x) · dx f(x)

Proof Put u = f(x) and y = lnu. Then y = ln f(x) and so

d

dxln[f(x)] =

dy

dx

= dy

du · du

dx chain rule

= d

du lnu · du dx = 1 u · du dx = 1

f(x) · d

16

Simple Form General Form

d dxx

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

16

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

16

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

16

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

dx cos[f(x)] = − sin[f(x)] ·

d

16

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

dx cos[f(x)] = − sin[f(x)] ·

d

dx f(x)

d

dx tan x = sec

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

dx cos[f(x)] = − sin[f(x)] ·

d

dx f(x)

d

dx tan x = sec

2 _x d

16

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

dx cos[f(x)] = − sin[f(x)] ·

d

dx f(x)

d

dx tan x = sec

2 _x d

dx tan[f(x)] = sec

2_{[f(x)] ·} d

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

dx cos[f(x)] = − sin[f(x)] ·

d

dx f(x)

d

dx tan x = sec

2 _x d

dx tan[f(x)] = sec

2_{[f(x)] ·} d

dx f(x)

d dxe

16

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

dx cos[f(x)] = − sin[f(x)] ·

d

dx f(x)

d

dx tan x = sec

2 _x d

dx tan[f(x)] = sec

2_{[f(x)] ·} d

dx f(x)

d dxe

x _{= e}x d

dxe

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

dx cos[f(x)] = − sin[f(x)] ·

d

dx f(x)

d

dx tan x = sec

2 _x d

dx tan[f(x)] = sec

2_{[f(x)] ·} d

dx f(x)

d dxe

x _{= e}x d

dxe

f(x) _{= e}f(x) _· d

16

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

dx cos[f(x)] = − sin[f(x)] ·

d

dx f(x)

d

dx tan x = sec

2 _x d

dx tan[f(x)] = sec

2_{[f(x)] ·} d

dx f(x)

d dxe

x _{= e}x d

dxe

f(x) _{= e}f(x) _· d

dx f(x)

d

dx ln x =

1

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

dx cos[f(x)] = − sin[f(x)] ·

d

dx f(x)

d

dx tan x = sec

2 _x d

dx tan[f(x)] = sec

2_{[f(x)] ·} d

dx f(x)

d dxe

x _{= e}x d

dxe

f(x) _{= e}f(x) _· d

dx f(x)

d

dx ln x =

1

x

d

16

Simple Form General Form

d dxx

r _{= rx}r−1 d

dx[f(x)]

r _{= r[f(x)]}r−1 _· d

dx f(x)

d

dx sin x = cos x

d

dx sin[f(x)] = cos[f(x)] ·

d

dx f(x)

d

dx cos x = −sin x

d

dx cos[f(x)] = − sin[f(x)] ·

d

dx f(x)

d

dx tan x = sec

2 _x d

dx tan[f(x)] = sec

2_{[f(x)] ·} d

dx f(x)

d dxe

x _{= e}x d

dxe

f(x) _{= e}f(x) _· d

dx f(x)

d

dx ln x =

1

x

d

dx ln[f(x)] =

1

f(x) ·

d

dx (1) y = sin(x2 + 1) (2) y = ex3+2

(3) y = ln(x4 − 3x + 2) (4) y = ex5+tanx5

(5) y = ln[sin2(2x + 3)] (6) y = 1

(x2 _{+ 3)}40 − 3e 4x+1

18 Example Find dy

dx for the following: (1) y = sin(x2 + 1)

dx (1) y = sin(x2 + 1)

Solution dy

dx =

d

dx sin(x

18 Example Find dy

dx for the following: (1) y = sin(x2 + 1)

Solution dy

dx =

d

dx sin(x

2 _{+ 1)}

= cos(x2 + 1) · d dx(x

dx (1) y = sin(x2 + 1)

Solution dy

dx =

d

dx sin(x

2 _{+ 1)}

= cos(x2 + 1) · d dx(x

2 _{+ 1)}

18 Example Find dy

dx for the following: (1) y = sin(x2 + 1)

Solution dy

dx =

d

dx sin(x

2 _{+ 1)}

= cos(x2 + 1) · d dx(x

2 _{+ 1)}

= 2xcos(x2 + 1) (2) y = ex3+2

dx (1) y = sin(x2 + 1)

Solution dy

dx =

d

dx sin(x

2 _{+ 1)}

= cos(x2 + 1) · d dx(x

2 _{+ 1)}

= 2xcos(x2 + 1) (2) y = ex3+2

Solution dy

dx =

d dxe

18 Example Find dy

dx for the following: (1) y = sin(x2 + 1)

Solution dy

dx =

d

dx sin(x

2 _{+ 1)}

= cos(x2 + 1) · d dx(x

2 _{+ 1)}

= 2xcos(x2 + 1) (2) y = ex3+2

Solution dy

dx =

d dxe

x3+2

= ex3+2 · d dx(x

dx (1) y = sin(x2 + 1)

Solution dy

dx =

d

dx sin(x

2 _{+ 1)}

= cos(x2 + 1) · d dx(x

2 _{+ 1)}

= 2xcos(x2 + 1) (2) y = ex3+2

Solution dy

dx =

d dxe

x3+2

= ex3+2 · d dx(x

3 _{+ 2)}

19 Example Find dy

dx for the following: (3) y = ln(x4 − 3x + 2)

(3) y = ln(x4 − 3x + 2) Solution dy

dx =

d

dx ln(x

19 Example Find dy

dx for the following: (3) y = ln(x4 − 3x + 2)

Solution dy

dx =

d

dx ln(x

4 _{− 3x + 2)}

= 1

x4 _{− 3x + 2} ·

d dx(x

(3) y = ln(x4 − 3x + 2) Solution dy

dx =

d

dx ln(x

4 _{− 3x + 2)}

= 1

x4 _{− 3x + 2} ·

d dx(x

4 _{− 3x + 2) =} 4x3 − 3

19 Example Find dy

dx for the following: (3) y = ln(x4 − 3x + 2)

Solution dy

dx =

d

dx ln(x

4 _{− 3x + 2)}

= 1

x4 _{− 3x + 2} ·

d dx(x

4 _{− 3x + 2) =} 4x3 − 3

x4 _{− 3x + 2}

(3) y = ln(x4 − 3x + 2) Solution dy

dx =

d

dx ln(x

4 _{− 3x + 2)}

= 1

x4 _{− 3x + 2} ·

d dx(x

4 _{− 3x + 2) =} 4x3 − 3

x4 _{− 3x + 2}

(4) y = ex5+tanx5 Solution dy

dx =

d dxe

19 Example Find dy

dx for the following: (3) y = ln(x4 − 3x + 2)

Solution dy

dx =

d

dx ln(x

4 _{− 3x + 2)}

= 1

x4 _{− 3x + 2} ·

d dx(x

4 _{− 3x + 2) =} 4x3 − 3

x4 _{− 3x + 2}

(4) y = ex5+tanx5 Solution dy

dx =

d dxe

x5+tanx5 _{= e}x5+tanx5 _· d dx(x

(3) y = ln(x4 − 3x + 2) Solution dy

dx =

d

dx ln(x

4 _{− 3x + 2)}

= 1

x4 _{− 3x + 2} ·

d dx(x

4 _{− 3x + 2) =} 4x3 − 3

x4 _{− 3x + 2}

(4) y = ex5+tanx5 Solution dy

dx =

d dxe

x5+tanx5 _{= e}x5+tanx5 _· d dx(x

5 _{+ tan x}5₎

= ex5+tanx5 · (5x4 + d

dx tan(x

19 Example Find dy

dx for the following: (3) y = ln(x4 − 3x + 2)

Solution dy

dx =

d

dx ln(x

4 _{− 3x + 2)}

= 1

x4 _{− 3x + 2} ·

d dx(x

4 _{− 3x + 2) =} 4x3 − 3

x4 _{− 3x + 2}

(4) y = ex5+tanx5 Solution dy

dx =

d dxe

x5+tanx5 _{= e}x5+tanx5 _· d dx(x

5 _{+ tan x}5₎

= ex5+tanx5 · (5x4 + d

dx tan(x

5_{)← do in your head)}

= ex5+tanx5 · (5x4 + sec2x5 · d dxx

(3) y = ln(x4 − 3x + 2) Solution dy

dx =

d

dx ln(x

4 _{− 3x + 2)}

= 1

x4 _{− 3x + 2} ·

d dx(x

4 _{− 3x + 2) =} 4x3 − 3

x4 _{− 3x + 2}

(4) y = ex5+tanx5 Solution dy

dx =

d dxe

x5+tanx5 _{= e}x5+tanx5 _· d dx(x

5 _{+ tan x}5₎

= ex5+tanx5 · (5x4 + d

dx tan(x

5_{)← do in your head)}

= ex5+tanx5 · (5x4 + sec2x5 · d dxx

5₎

19 Example Find dy

dx for the following: (3) y = ln(x4 − 3x + 2)

Solution dy

dx =

d

dx ln(x

4 _{− 3x + 2)}

= 1

x4 _{− 3x + 2} ·

d dx(x

4 _{− 3x + 2) =} 4x3 − 3

x4 _{− 3x + 2}

(4) y = ex5+tanx5 Solution dy

dx =

d dxe

x5+tanx5 _{= e}x5+tanx5 _· d dx(x

5 _{+ tan x}5₎

= ex5+tanx5 · (5x4 + d

dx tan(x

5_{)← do in your head)}

= ex5+tanx5 · (5x4 + sec2x5 · d dxx

5₎