Calculus Cheat Sheet Derivatives - Pauls Online Math Notes

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Calculus Cheat SheetChain Rule VariantsThe chain rule applied to some specific functions.dnn-1d1. ( éf ( x)ù ) = néf ( x) ù f¢( x)dxë û ë û5. ( coséf ( x)) f ( x) sin f ( x)dxë ùû =- ¢ éë ùûd f ( x)f ( x)d22. ( e ) = f¢( x)e6. ( tanéf ( x)) f ( x) sec f ( x)dxdxë ùû = ¢ éë ùûdf¢( x)d3. ( ln éf( x)ù)=7. secdxë û[ ] = ¢ sec tanf ( x)dxdd 1f ( x)4. ( sinéf ( x)ù) = f¢( x) cos éf ( x)ù8. ( tan- éf( x))¢2dxë û ë û dxë ùû=1+éëf ( x)ùû( f ( x) ) f ( x) [ f ( x) ] [ f ( x)]Higher Order DerivativesThe Second Derivative is denoted asThe n th Derivative is denoted as2n( 2( ) ) d f( nf¢¢ x = f ( x)= and is defined as) d ff2( x)= and is defined asndxdxf¢¢ ( x) = f¢( x )¢nn- 1, i.e. the derivative of the¢f x = f x , i.e. the derivative of( )¢ .first derivative, f ( x)( )( )(()( ))the (n-1) st derivative, f( n-1)( x)Implicit Differentiation+ xy = sin y + 11xy = y x here, so products/quotients of x and y2x-9y3 2Find y¢ if e ( ) . Remember ( )will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is todifferentiate as normal and every time you differentiate a y you tack on a y¢ (from the chain rule).After differentiating solve for y¢.e( y¢ )exy xyy¢ ( y)y¢( )-x- ey- ( ) ¢ = -x-ey-2x-9y2 2 32- 9 + 3 + 2 = cos + 11- y¢ + xy + xyy¢ = y y¢ + Þ y¢=2x-9y 2x-9y2 2 32e9 3 2 cos 11( )2xy 9 cos y y 11 2 3xy3 2 9 2 9 2 2Critical Pointsx c= is a critical point of f ( )1. f¢ ( c) = 0 or 2. f ( c)Increasing/Decreasing – Concave Up/Concave Down¢ doesn’t exist.x provided eitherIncreasing/Decreasing1. If f¢ ( x) > 0 for all x in an interval I thenf ( x ) is increasing on the interval I.¢ < for all x in an interval I then2. If f ( x) 0f ( x ) is decreasing on the interval I.¢ = for all x in an interval I then3. If f ( x) 0f ( x ) is constant on the interval I..2x-9y2 211-2e-3xy3 2x-9y2xy-9e-cos( y)Concave Up/Concave Down1. If f¢¢ ( x) > 0 for all x in an interval I thenf ( x ) is concave up on the interval I.¢¢ < for all x in an interval I then2. If f ( x) 0f ( x ) is concave down on the interval I.Inflection Pointsx c= is a inflection point of f ( )concavity changes at x= c.x if theVisit http://tutorial.math.lamar.edu for a complete set of Calculus notes.© 2005 Paul Dawkins
Absolute Extrema1. x c= is an absolute maximum of f ( x )if f ( c) ³ f ( x)for all x in the domain.= is an absolute minimum of f ( x )2. x cif f ( c) £ f ( x)for all x in the domain.Fermat’s Theoremf x has a relative (or local) extrema atIf ( )x= c, then x c= is a critical point of ( )Calculus Cheat Sheetf x .Extreme Value Theoremf x is continuous on the closed intervalIf ( )[ ab , ] then there exist numbers c and d so that,1. a£ cd , £ b, 2. f ( c ) is the abs. max. in[ ab , ] , 3. ( )f d is the abs. min. in [ , ]ab .Finding Absolute ExtremaTo find the absolute extrema of the continuousab , use thefunction f ( x ) on the interval [ ]following process.1. Find all critical points of f ( x ) in [ ab , ] .2. Evaluate f ( x ) at all points found in Step 1.3. Evaluate f ( a ) and f ( b ) .4. Identify the abs. max. (largest functionvalue) and the abs. min.(smallest functionvalue) from the evaluations in Steps 2 & 3.ExtremaRelative (local) Extrema1. x= c is a relative (or local) maximum off c ³ f x for all x near c.f ( x ) if ( ) ( )2. x= c is a relative (or local) minimum off c £ f x for all x near c.f ( x ) if ( ) ( )1 st Derivative TestIf x c= is a critical point of ( )1. a rel. max. of f ( x ) if f ( x) 0f x then x= c is¢ > to the leftof x= c and f¢ ( x) < 0 to the right of x= c.2. a rel. min. of f ( x ) if f ( x) 0¢ < to the leftof x= cand f¢ ( x) > 0to the right of x= c.¢ is3. not a relative extrema of f ( x ) if f ( x)the same sign on both sides of x2 nd Derivative TestIf x cMean Value Theorem,= c.= is a critical point of f ( x ) such thatf¢ ( c) = 0 then x=c1. is a relative maximum of f ( x ) if f ( c) 02. is a relative minimum of f ( x ) if f ( c) 03. may be a relative maximum, relativeminimum, or neither if f¢¢ ( c) = 0 .Finding Relative Extrema and/orClassify Critical Points1. Find all critical points of f ( x ) .2. Use the 1 st derivative test or the 2 ndderivative test on each critical point.¢¢ < .¢¢ > .If f ( x ) is continuous on the closed interval [ ab ] and differentiable on the open interval ( ab , )f ( b) - f ( a)then there is a number a< c< b such that f¢ ( c)=.b-aNewton’s MethodIf xnis the n th guess for the root/solution of f ( x ) = 0 then (n+1) st guess isprovided f¢ ( x n ) exists.xxf= - n+ 1 n¢f( xn)( x )nVisit http://tutorial.math.lamar.edu for a complete set of Calculus notes.© 2005 Paul Dawkins
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Calculus Cheat Sheet Derivatives - Pauls Online Math Notes

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