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

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

Examples Find the derivative: 1. f(x) = 5x + 4x3 +3sinx f’(x) = 5 + 12x2 + 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)

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)

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

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)

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

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

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 = 75.707

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)