MATH 111 Lecture Notes - Lecture 13: Indeterminate Form, Inverse Trigonometric Functions, Minute And Second Of Arc
9 Jun 2015
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1476- lecture 13 - derivatives of logarithms and l"hospital"s rule. Take the derivative of both sides (d/dx) Dy/dx (arccos x) = - 1/ (1-x2) Dy/dx (arccsc x) = - 1/ |x| (x2-1) Dy/dx (arccot x) = - 1/ (1+x2) D/dx (log ax) = 1/ (x lna) Because of the chain rule: ln (-x)" = 1/ (-x) *-1= 1/x. Both of these can be combined with the chain rule: (log au)" = 1/ (u lna) * du. Logarithmic differentiation take the natural logarithm of both sides of the equation and simplify as much as possible. 1: differentiate implicitly with respect to x, solve for dy/dx. Ex: y= [ x3/4 (x2+1)] / (3x+2)5. Take the natural logarithm of both sides. Ln y = ln x3/4 + ln (x2+1) - ln (3x+2)5. Simplify as much as possible using the laws of logarithms. Ln y = ( ) ln x + ln (x2+1) - 5 ln (3x+2)
