Solution to Review Problems for Midterm II Midterm II: Monday, October 18 in class Topics: 31-3 (except 34) 1 Use te definition of derivative f f(x+) f(x) (x) lim 0 to find te derivative of te functions (a) f(x) 2x + 3 (b) f(x) 1 2x+3 Solution: (a) First, let us find te expression f(x + ) f(x) 2(x + ) + 3 2x + 3 2x + 2 + 3 2x + 3 Now f(x+) f(x) 2x+2+3 2x+3 2x+2+3 2x+3 2x+2+3+ 2x+3 2x+2+3+ 2x+3 ( 2x+2+3) 2 ( 2x+3) 2 ( 2x+2+3+ 2x+3) 2x+2+3 (2x+3) ( 2x+2+3+ 2x+3) Tus f (x) lim 0 f(x+) f(x) lim 0 2 2x+2+3+ 2x+3 2 2x+3+ 2x+3 2 2 2x+3 1 2x+3 (b) First, let us find te expression 1 f(x + ) f(x) 1 1 1 2(x+)+3 2x+3 2x+2+3 2 ( 2x+2+3+ 2 2x+3) ( 2x+2+3+ 2x+3) 2x+3 2x+3 2x+2+3 2x+3 (2x+2+3) 2x+3 2x 2 3 (2x+2+3)(2x+3) (2x+2+3)(2x+3) (2x+2+3)(2x+3) (2x+2+3)(2x+3) Now f(x+) f(x) 2) (2x+2+3)(2x+3) 2 (2x+2+3)(2x+3) 2 f (x) lim 0 f(x+) f(x) lim 0 2 2 (2x+3)(2x+3) 2 (2x+3) 2 (2x+2+3)(2x+3) (2x+2+3)(2x+3) 2 (2x+2+3)(2x+3) Tus 2 Find te derivative of te following functions and simplify your answers (a) 12x 3 + 4x 2 x 2 (b) (1 + 4x)e 4x (c) (sec(x) + tan(x)) 3 (d) x cos(x) 6x sin(x) 6 cos(x) (e) ( cos(x) 1+sin(x) ) Solution: (a) First, we rewrite 12x 3 x 2 + 4x 2 12x 3 x 2 + 4x 2 (12x 3 +4x 2 x 2 ) (12x 3 x 2 +4x 2 60x 4 + 6 x 3 8x ) 12 x 4 3 ( 2)x 3 +4 ( 2 So ) x (b) Applying te product rule, we ave ((1 + 4x)e 4x ) (1 + 4x) e 4x + (1 + 4x)(e 4x ) 4e 4x + (1 + 4x)e 4x ( 4x) 4e 4x + (1 + 4x)e 4x ( 4) 4e 4x 4e 4x 16xe 4x 16xe 4x (c) [(sec(x) + tan(x)) 3 ] 3(sec(x) + tan(x)) 2 (sec(x) + tan(x)) 3(sec(x) + tan(x)) 2 (sec(x) tan(x) + sec 2 (x)) 3(sec(x) + tan(x)) 2 sec(x) (sec(x) + sec(x)) 3 sec(x)(sec(x) + tan(x)) 3 (d)(x cos(x) 6x sin(x) 6 cos(x)) (x cos(x)) (6x sin(x)) (6 cos(x)) x 4 cos(x) + x ( sin(x)) 6 sin(x) 6x cos(x) 6( sin(x)) x 4 cos(x) x sin(x) 6 sin(x) 6x cos(x) + 6 sin(x) x 4 cos(x) x sin(x) 6x cos(x) MATH 180: page 1 of 6
Solution to Review Problems for Midterm II MATH 180: page 2 of 6 (e) [( cos(x) 1+sin(x) ) ] ( cos(x) 1+sin(x) )4 ( cos(x) 1+sin(x) ) ( cos(x) ( cos(x) 1+sin(x) )4 ( ( sin(x))(1+sin(x)) cos(x) cos(x) (1+sin(x)) 2 ( cos(x) 1+sin(x) )4 ( sin(x) 1 (1+sin(x)) 2 ) ( cos(x) 1+sin(x) )4 ( (cos(x)) (1+sin(x)) cos(x)(1+sin(x)) ) (1+sin(x)) 2 ) ( cos(x) 1+sin(x) )4 ( sin(x) sin2 (x) cos 2 (x) ) (1+sin(x)) 2 1+sin(x) )4 1 ( ) cos4 (x) (1+sin(x)) (1+sin(x)) 3 Find te derivative of te following functions You don t ave to simplify your answer (a) (2x + 1) 3 (1 + e 2x ) (b) (2x+1)3 (c) tan(sin(xe x )) (d) cot 6 ( 2) (1+e 2x ) t (e) (f) e 4x sec(x2) (g) sin 3 (2t) cos 3 (2t) () x 3 tan 3 ((1 + x 2 ) 2 ) 2 +e x2 (i) ex2 csc(3x) x 2 (j) x 4 e 3x cos(x) (k) sin (2x) (1+x 2 ) 2 x Solution:(a) [(2x + 1) 3 (1 + e 2x ) ] {}}{}{{} [(2x + 1) 3 ] }{{} (1 + e2x ) + (2x + 1) 3 [(1 + e 2x ) ] product rule x cos3 (2x) 3 (l) 1 + t cos(t 2 ) 2t3 3 sin(t2 ) {}}{ 3(2x + 1) 2 (2x + 1) }{{} (1 + e2x ) + (2x + 1) 3 (1 + e 2x ) 4 (1 + e 2x ) 3(2x + 1) 2 2 (1 + e 2x ) + (2x + 1) 3 (1 + e 2x ) 4 e 2x (2x) 6(2x + 1) 2 (1 + e 2x ) + (2x + 1) 3 (1 + e 2x ) 4 e 2x 2 6(2x + 1) 2 (1 + e 2x ) + 10(2x + 1) 3 (1 + e 2x ) 4 (b) First, we can rewrite (2x+1)3 (2x + 1) 3 (1 + e 2x ) (1+e 2x ) [ (2x+1)3 ] [(2x + 1) 3 (1 + e 2x ) ] (1+e 2x ) {}}{}{{} [(2x + 1) 3 ] }{{} (1 + e2x ) + (2x + 1) 3 [(1 + e 2x ) ] product rule {}}{ 3(2x + 1) 2 (2x + 1) }{{} (1 + e2x ) + (2x + 1) 3 ( ) (1 + e 2x ) 6 (1 + e 2x ) 3(2x + 1) 2 2 (1 + e 2x ) + (2x + 1) 3 ( ) (1 + e 2x ) 6 e 2x (2x) 6(2x + 1) 2 (1 + e 2x ) + (2x + 1) 3 ( ) (1 + e 2x ) 6 e 2x 2 6(2x + 1) 2 (1 + e 2x ) 10(2x + 1) 3 (1 + e 2x ) 6 (c) [tan(sin(xe x ))] sec 2 (sin(xe x ))[sin(xe x )] sec 2 (sin(xe x )) cos(xe x )(xe x ) sec 2 (sin(xe x )) cos(xe x )(e x + xe x ) (d) Note tat cot 6 ( 2 t ) (cot(2t 1 )) 6 So [cot 6 ( 2 t )] [(cot(2t 1 )) 6 ] 6(cot(2t 1 )) [cot(2t 1 )] 6(cot(2t 1 )) [ csc 2 (2t 1 )(2t 1 ) 6(cot(2t 1 )) [ csc 2 (2t 1 )]( 2t 2 ) (e) We can rewrite 4 x 2 +e x2 (x 2 +e x2 ) 1 4 [(x 2 + e x2 ) 1 4 ] [(x 2 + e x2 ) 1 4 ] (x 2 + e x2 ) 1 4 So ( 4x 2 +e x2 )
MATH 180: page 3 of 6 Solution to Review Problems for Midterm II ( 1 4 ) (x2 + e x2 ) 4 (x 2 + e x2 ) ( 1 4 ) (x2 + e x2 ) 4 (2x + e x2 (x 2 ) ) ( 1) 4 (x2 + e x2 ) 4 (2x + e x2 (2x)) (f) (e sec(x2) ) e sec(x2) (sec(x 2 )) }{{} esec(x2) sec(x 2 ) tan(x 2 )(x 2 ) }{{} e sec(x2) sec(x 2 ) tan(x 2 ) (2x) (g) (sin 3 (2t) cos 3 (2t)) (sin 3 (2t)) cos 3 (2t) + sin 3 (2t)(cos 3 (2t)) 3(sin 2 (2t)) (sin(2t)) cos 3 (2t) + sin 3 (2t) 3 (cos 2 (2t))(cos(2t)) 3(sin 2 (2t)) cos(2t) 2 cos 3 (2t) + sin 3 (2t) 3 (cos 2 (2t))( sin(2t)) 2 () [x 3 tan 3 ((1 + x 2 ) 2 )] (x 3 ) tan 3 ((1 + x 2 ) 2 ) + x 3 [tan 3 ((1 + x 2 ) 2 )] 3x 2 tan 3 ((1 + x 2 ) 2 ) + x 3 3 tan 2 ((1 + x 2 ) 2 )[tan((1 + x 2 ) 2 )] 3x 2 tan 3 ((1 + x 2 ) 2 ) + x 3 3 tan 2 ((1 + x 2 ) 2 )[sec 2 ((1 + x 2 ) 2 )][(1 + x 2 ) 2 ] 3x 2 tan 3 ((1 + x 2 ) 2 ) + x 3 3 tan 2 ((1 + x 2 ) 2 )[sec 2 ((1 + x 2 ) 2 )] 2 (1 + x 2 ) (2x) (i)note tat ex2 csc(3x) x 2 (1+x 2 ) 2 (e x2 csc(3x) x 2 )(1 + x 2 ) 2 So ( ex2 csc(3x) x 2 ) [(e x2 csc(3x) x 2 )(1 + x 2 ) 2 ] (1+x 2 ) 2 [(e x2 csc(3x) x 2 )] (1 + x 2 ) 2 + (e x2 csc(3x) x 2 )[(1 + x 2 ) 2 ] ([e x2 csc(3x)] 2x)(1 + x 2 ) 2 + (e x2 csc(3x) x 2 ) ( 2)[(1 + x 2 ) 3 ](1 + x 2 ) ((e x2 ) csc(3x) + e x2 (csc(3x)) 2x)(1 + x 2 ) 2 + (e x2 csc(3x) x 2 ) ( 2)[(1 + x 2 ) 3 ] (2x) ([e x2 (2x) csc(3x) + e x2 ( csc(3x) cot(3x)) 3] 2x)(1 + x 2 ) 2 + (e x2 csc(3x) x 2 ) ( 2)[(1 + x 2 ) 3 ](2x) (j) (x 4 e 3x cos(x)) (x 4 e 3x ) cos(x) + x 4 e 3x (cos(x)) [(x 4 ) e 3x + x 4 (e 3x ) ] cos(x) + x 4 e 3x ( sin(x)) [4x 3 e 3x + x 4 e 3x ( 3)] cos(x) + x 4 e 3x ( sin(x)) (k) Note tat sin (2x) x So ( sin (2x) x x cos3 (2x) 3 sin (2x)x 1 1 3 x cos3 (2x) x cos3 (2x) 3 ) (sin (2x)x 1 ) 1 3 (x cos3 (2x)) (sin (2x)) x 1 + sin (2x)(x 1 ) 1 3 (x cos 3 (2x) + x(cos 3 (2x)) ) ( ) (sin 6 (2x))(sin(2x)) x 1 +sin (2x) ( 1)x 2 1 3 (cos3 (2x)+x 3(cos 2 (2x))(cos(2x)) ) ( ) (sin 6 (2x))(cos(2x)) 2 x 1 + sin (2x) ( 1)x 2 1 3 (cos3 (2x) + x 3(cos 2 (2x))( 2 sin(2x)) 2) (l) Note tat So ( 1 + t cos(t 2 ) 2t3 3 sin(t2 ) [1 + t cos(t 2 ) 2t3 3 sin(t2 )] 1 2 1 + t cos(t 2 ) 2t3 3 sin(t2 )) ([1 + t cos(t 2 ) 2t3 3 sin(t2 )] 1 2 ) 1[1 + t 2 cos(t2 ) 2t3 3 sin(t2 )] 1 2 [1 + t cos(t 2 ) 2t3 3 sin(t2 )] 1[1 + t 2 cos(t2 ) 2t3 3 sin(t2 )] 1 2 [t cos(t 2 )+t(cos(t 2 )) ( 2t3 3 ) sin(t 2 ) 2t3 3 (sin(t2 )) ]
Solution to Review Problems for Midterm II MATH 180: page 4 of 6 1[1 + t 2 cos(t2 ) 2t3 3 sin(t2 )] 1 2 [cos(t 2 ) + t( sin(t 2 )) (2t) (2t 2 ) sin(t 2 ) 2t 3 3 (cos(t2 )) 2t] 4 Find te first derivative (y ) and second derivative (y ) of te following functions (a) y (6 + 4 x ) (b) y x 3 e 3x Solution: (a)note tat y (6 + 4 x ) (6 + 4x 1 ) So y [(6 + 4x 1 ) ] (6 + 4x 1 ) 4 (6 + 4x 1 ) (6 + 4x 1 ) 4 ( 4x 2 ) 20(6 + 4x 1 ) 4 x 2 y [ 20(6 + 4x 1 ) 4 x 2 ] 20[(6 + 4x 1 ) 4 ] x 2 20(6 + 4x 1 ) 4 (x 2 ) 20 4 (6 + 4x 1 ) 3 ((6 + 4x 1 )) x 2 20(6 + 4x 1 ) 4 ( 2x 3 ) 20 4 (6 + 4x 1 ) 3 ( 4x 2 ) x 2 + 40(6 + 4x 1 ) 4 x 3 320 (6 + 4x 1 ) 3 x 4 + 40(6 + 4x 1 ) 4 x 3 (b) y (x 3 e 3x ) (x 3 ) e 3x + x 3 (e 3x ) 3x 2 e 3x + x 3 e 3x 3 3x 2 e 3x + 3x 3 e 3x (3x 2 + 3x 3 )e 3x y [(3x 2 + 3x 3 )e 3x ] (3x 2 + 3x 3 ) e 3x + (3x 2 + 3x 3 )(e 3x ) (6x + 9x 2 )e 3x + (3x 2 + 3x 3 )e 3x 3 (6x + 9x 2 )e 3x + (9x 2 + 9x 3 )e 3x (6x + 9x 2 + 9x 2 + 9x 3 )e 3x (6x + 18x 2 + 9x 3 )e 3x Use implicit differentiation to find dy dx (a) 2xy y 2 x (b) x 3 + 3x 2 y + y 3 8 (c) x+y x y x2 + y 2 (d) cos(xy) + x y (e) e xy sin(x + y) In implicit differentiation, Suppose y y(x) ten (f(y)) f (y)y Solution:(a) Differentiating 2xy y 2 x, we get 2(xy) (y 2 ) x 2x y+2xy 2yy 1 2y+2xy 2yy 1 Now we ave 2xy 2yy 1 2y y (2x 2y) 1 2y y 1 2y 2x 2y (b) Differentiating te equation, we get (x 3 + 3x 2 y + y 3 ) (8) 3x 2 + 3(x 2 ) y + 3x 2 y + 3y 2 y 0 and 3x 2 + 6xy + 2x 2 y + 3y 2 y 0 Tis implies tat 2x 2 y + 3y 2 y 3x 2 6xy, y (2x 2 + 3y 2 ) 3x 2 6xy and y 3x2 6xy 2x 2 +3y 2 (c) Note tat x+y x y x2 + y 2 is te same as (x + y) (x y)(x 2 y 2 ) Differentiating te equation, we get (x + y) [(x y)(x 2 y 2 )] 1 + y (x y) (x 2 y 2 ) + (x y)(x 2 y 2 ) 1 + y (1 y )(x 2 y 2 ) + (x y)(2x 2yy ) 1 + y x 2 y 2 y (x 2 y 2 ) + 2x(x y) 2y(x y)y y + y (x 2 y 2 ) + 2y(x y)y x 2 y 2 + 2x 2 2xy 1 3x 2 y 2 2xy 1 y (1 + x 2 y 2 + 2yx 2y 2 ) 3x 2 y 2 2xy 1 and y (1 + x 2 3y 2 + 2yx) 3x 2 y 2 2xy 1 Tis gives y 3x2 y 2 2xy 1 1+x 2 3y 2 +2yx
MATH 180: page of 6 Solution to Review Problems for Midterm II (d) [cos(xy) + x ] (y ) sin(xy)(xy) + x 4 y 4 y sin(xy)(y + xy ) + x 4 y 4 y sin(xy)y sin(xy)xy + x 4 y 4 y sin(xy)xy y 4 y sin(xy)y x 4 y ( sin(xy)x y 4 ) sin(xy)y x 4 y sin(xy)y x4 sin(xy)x y 4 (e) (e xy ) (sin(x + y)) e xy (xy) cos(x + y)(x + y) e xy (y+xy ) cos(x+y)(1+y ) e xy y+e xy xy cos(x+y)+ cos(x+y)y cos(x + y)y + e xy xy cos(x + y) e xy y y ( cos(x + y) + e xy x) cos(x + y) e xy y y cos(x+y) exy y cos(x+y)+e xy x 6 Sow tat (1, 2) lie on te curve 2x 3 + 2y 3 9xy 0 Ten find te te tangent and normal to te curve at (1, 2) Solution: Plugging (1, 2) to te equation 2x 3 + 2y 3 9xy, we get 2 1 3 + 2 2 3 9 1 2 2 + 2 8 18 2 + 16 18 0 Tis means tat (1, 2) lie on te curve 2x 3 + 2y 3 9xy 0 Next we find y by implicit differentiation Differentiating 2x 3 + 2y 3 9xy 0, we get 2(x 3 ) + 2(y 3 ) 9(xy) 0 6x 2 + 6y 2 y 9y 9xy 0 6y 2 y 9xy 6x 2 + 9y y (6y 2 9x) 6x 2 + 9y y 6x2 +9y 6y 2 9x At (1, 2), we ave y (1) 6 12 +9 2 6+18 12 4 So te slope of 6 2 2 9 1 24 9 1 te tangent line is m 4 and te point is (1, 2) By te point slope formula, we ave y 2 4 (x 1) From te slope of te tangent line, we know tat te slope of te normal line is m 4 So te equation of te normal line is y 2 (x 1) 4 Find te normal to te curve xy + 2x y 0 tat are parallel to te line x + 2y 0 Solution: First we find y by implicit differentiation Differentiating xy + 2x y 0, we get (xy) + 2(x) (y) 0 y + xy + 2 y 0 xy y y 2 So te slope of te tangent line at (x, y) is y 2 From ere, we know tat te slope of te normal line is x 1 x 1 x 1 Te equation x + 2y 0 can be rewritten as 2y x and y 2 y+2 y 1x So te slope is m 1 At (x, y), te slope of te normal to 2 2 te curve xy + 2x y 0 tat are parallel to te line x + 2y 0 must ave slope 1 x 1 Tis implies tat 1, 2x 2 y 2 and y 2x 2 y+2 2 y (x 1) y 2 y y 2 x 1
Solution to Review Problems for Midterm II MATH 180: page 6 of 6 Plugging y 2x into te equation of te curve xy + 2x y 0, we get x( 2x) + 2x ( 2x) 0 2x 2 + 2x + 2x 0 2x 2 + 4x 0 2x(x 2) 0 x 0or x 2 Recall tat y 2x Tis implies tat y 0 or y 4 So te point is (0, 0) and te slope is (2, 4) Recall te slope of te normal line is 1 So te normal line parallel to x + 2y 0 2 are y 1x (point(0, 0))and y + 4 1 (x 2) (point(2, 4)) 2 2