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M2Review

# M2Review - Midterm 2 Review Material from the previous...

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Midterm 2 Review Material from the previous midterm The midterm is not cumulative but you are expected to know the following important facts: d dx (1) = 0 d dx ( x a ) = ax a - 1 d dx ( f ( x ) + g ( x ) ) = f 0 ( x ) + g 0 ( x ) d dx ( kf ( x ) ) = kf 0 ( x ) d dx ( f ( x ) g ( x ) ) = f 0 ( x ) g ( x ) + f ( x ) g 0 ( x ) This midterm will not test on limits, tangent lines, Squeeze Theorem, or other facts from the first midterm (though they may make a comeback for the final!) Derivatives of trigonometric functions Using the limit of sin( h ) /h as h 0 and some basic trigonometric identities we were able to find deriva- tives for the trigonometric functions. d dx ( sin x ) = cos x d dx ( cos x ) = - sin x d dx ( tan x ) = ( sec x ) 2 d dx ( sec x ) = sec x tan x The important thing to remember is to get the signs on the derivative of sine and cosine correct. Since we can now be tested on these derivatives it is also help- ful to know some of the values of the trigonometric functions. θ 0 1 6 π 1 4 π 1 3 π 1 2 π sin θ 0 1 2 1 2 2 1 2 3 1 cos θ 1 1 2 3 1 2 2 1 2 0 tan θ 0 1 3 3 1 3 DNE Chain rule The chain rule deals with the situation of how to take the derivative of a function inside of another func- tion. Often a hint that we will use the chain rule is parentheses “(” and “)” or if when reading it out loud we use the word “of”. For example sin( x 2 ) (read “sine of x squared”) is the function x 2 inside the sine func- tion. d dx ( f ( g ( x ) )) = f 0 ( g ( x ) ) g 0 ( x ) Common mistakes when using the chain rule include forgetting to multiply by g 0 ( x ) or putting f 0 ( g 0 ( x )). So for example d dx ( sin ( x 2 ) ) = cos( x 2 ) · 2 x and not cos(2 x ) or cos( x 2 ). Also make sure you have clearly identified what the inside and outside func- tions are, i.e., sin( x 2 ) is a very different function than (sin x ) 2 . An important application of the chain rule is the power rule, which is the special case when f ( x ) = x a . d dx (( g ( x ) ) a ) = a ( g ( x ) ) a - 1 g 0 ( x ) . Implicit differentiation An explicit function is when you have y = stuff with x . An implicit function is when we are given a relation- ship involving x and y but we do not (or can not) write y as a simple function of x . For example, yx 2 + 3 x cos y = sin x + y gives a relationship between x and y , and for a given value of x we can solve for what possible value(s) y can be. So even though y is not given as a function of x we still can think of it as a function of x (simply being the case that we don’t know exactly which function it is). So to take the derivative we start with the equation defining the implicit relationship and take the deriva- tive of both sides with respect to x . There are a few things to observe 1. If we take the derivative of two functions which are equal the derivatives are still equal. 2. We treat y as a function of x , in particular when it comes to expressions involving y we use the chain rule to take the derivative and when we finally get to the point where we need to write down the derivative of y we put dy/dx or y 0 (both mean the derivative of y , how easy is that!).

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