1. By: Kelley Borgard Block 4A
Topic 9
By: Kelley Borgard
Block 4A

2. Theorems 2.6 and 2.9 2.6: Derivatives of Sine and Cosine Functions
d/dx [sin x] = cos x
d/dx [cos x] = -sin x
2.9: Derivatives of Trigonometric Functions
d/dx [tan x] = sec2x
d/dx [sec x] = sec x tan x
d/dx [cot x] = -csc2x
d/dx [csc x] = -csc x cot x

3. Examples Find the derivative: 1. f(x) = 5x + 4x3 +3sinx
f’(x) = x2 + 3cosx
2. h(x)=3x cos(4x+2) (product rule)
f(x) = 3x
g(x) = cos(4x+2)
f’(x) = 3
g’(x) = -sin(4x+2)(4)
= -4sin(4x+2)
h’(x) = f(x)g’(x) + g(x)f’(x)
= 3x[-4sin(4x+ 2)] + cos(4x +2)(3)
= -12x sin (4x + 2) + 3cos (4x +2)

4. Derivatives of Logarithmic Functions
d/dx lnx = 1/x
Example:
d/dx 5lnx= 5(1/x) = 5/x
d/dx ln(x2+4)
= (1/x2+4) (2x) = 2x/(x2+4)

5. Natural Log Ln (1) = 0 Ln (ab) = Ln(a) + Ln(b) Ln (an) = nLna
* a and b are positive and n is rational

6. Examples Expand a=● b=● n=● ln x(x2+1)2 = lnx + ln(x2+1)2 (product)
ln (x2+1)2 = 2ln(x2+1) (power)
ln x(x2+1)2/√(2x3-1) =ln[x(x2+1)2]-ln√(2x3-1) (quotient)

7. Using Log Properties for Differentiation
Ex.
Ln x(x2+1)2/√(2x3-1)
Solution:
Expand with log properties
Differentiate
=ln[x(x2+1)2] –ln√(2x3-1)
=lnx +ln(x2+1)2-ln(2x3-1)1/2
=lnx + 2ln(x2+1) -1/2 ln(2x3-1)
f’(x) =1/x + 2(1/x2+1)(2x) - ½(1/2x3-1)(6x2)
= 1/x + 4x/x2+1 -3x2/2x3-1

8. Derivative of ex Definition of the Natural Exponential Function:
If f(x) = lnx,
Then f-1(x) = ex
Inverse Relationship
ln(ex) = x
elnx = x

9. Examples a. 7=ex+1 ln(7) = ln(ex+1) ln(7) =x+1 x = ln(7) -1 = 0.946
b. ln(2x+3)=5
eln(2x-3) = e5
2x-3 = e5
2x = e5 +3
x = (e5 +3)/2
x =

10. The Derivative of ex d/dx ex = ex Example for Chain Rule: d/dx ex2+2
=ex2+2(2x)
=2xex2+2
Example for Product Rule:
d/dx 5x2ex^2+2
f(x) = 5x2
g(x)= ex^2+2
f’(x) = 10x
g’(x) = 2xex^2+2
f’(x) = 5x2(2xex^2+2) + (ex^2+2)10x
= 10x3ex^2+2+10xex^2+2
= 10xex^2+2(x2+1)