local/tmp/image-0001...3. Either integrate by parts or use Fubini's theorem: r—l x P(X > x) dx r f...

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Transcript of local/tmp/image-0001...3. Either integrate by parts or use Fubini's theorem: r—l x P(X > x) dx r f...

Page 1: local/tmp/image-0001...3. Either integrate by parts or use Fubini's theorem: r—l x P(X > x) dx r f (y) dy clx r—l rx dx dy — yr f (y) dy. An alternative proof is as follows.
Page 2: local/tmp/image-0001...3. Either integrate by parts or use Fubini's theorem: r—l x P(X > x) dx r f (y) dy clx r—l rx dx dy — yr f (y) dy. An alternative proof is as follows.

CALCULUS · y = 2 when x — 1 when x — 3C, - —9C1 sin3x — 9C2 cos 31 ... the points given in the table. 2 4 14 o 4 o 4 6 dy/dx — X COS dy/dx o 4 4 o 209 . ... Consider Example

CALCULUS · y = 2 when x — 1 when x — 3C, - —9C1 sin3x — 9C2 cos 31 ... the points given in the table. 2 4 14 o 4 o 4 6 dy/dx — X COS dy/dx o 4 4 o 209 . ... Consider Example

 · Exercise 5.3 2x + 3y = sin x Differentiating both sides with respect to x, we get — sin x dx dx dx dy cosx — 2 dx 2x + = sin y dy dx cos x Question 2: 2x + = sin y Answer

 · Exercise 5.3 2x + 3y = sin x Differentiating both sides with respect to x, we get — sin x dx dx dx dy cosx — 2 dx 2x + = sin y dy dx cos x Question 2: 2x + = sin y Answer

DERIVATIVES 3. Summary f(x) ≈ f(a) + f’(a)(x – a) L(x) = f(a) + f’(a)(x – a) ∆y = f(x + ∆x) – f(x) dx = ∆x dy = f’(x)dx ∆ y≈ dy.

DERIVATIVES 3. Summary f(x) ≈ f(a) + f’(a)(x – a) L(x) = f(a) + f’(a)(x – a) ∆y = f(x + ∆x) – f(x) dx = ∆x dy = f’(x)dx ∆ y≈ dy.

Ordinary Differential Equations - Numerical FactoryOrdinary Differential Equations • Analytical Solution: (only in few cases) • Separate variables: dy dx x dy x dx y x C ³ ³

Ordinary Differential Equations - Numerical FactoryOrdinary Differential Equations • Analytical Solution: (only in few cases) • Separate variables: dy dx x dy x dx y x C ³ ³

Computer Graphics 7 - Rasterisation · Testing for the side of a line I Assumethelineisbetween(x l,y l) and(x r,y r) I Theslopeofthelinewillbe dy dx wheredx = x r −x l and dy =

Computer Graphics 7 - Rasterisation · Testing for the side of a line I Assumethelineisbetween(x l,y l) and(x r,y r) I Theslopeofthelinewillbe dy dx wheredx = x r −x l and dy =

(-x-)zliu/math3280/1.5,1.6,2.1,2.3,3.1...dy 2 x --4x y+2yln y=O. dx 59. Solve the differ·~ntial equation dy X-y-I dx x + y + 3 by finding h and k so that the substitutions x = u +

(-x-)zliu/math3280/1.5,1.6,2.1,2.3,3.1...dy 2 x --4x y+2yln y=O. dx 59. Solve the differ·~ntial equation dy X-y-I dx x + y + 3 by finding h and k so that the substitutions x = u +

Extension 1 Mathematics 3U - Sydney...d) Find dy if y = e 2x cos-1 x dx e) Find dy at t = ::. if x = 2 cos t and y = 2 sin t dx 4 2x f) Solve - ~ x + 4 x-3 g) Use the substitution

Extension 1 Mathematics 3U - Sydney...d) Find dy if y = e 2x cos-1 x dx e) Find dy at t = ::. if x = 2 cos t and y = 2 sin t dx 4 2x f) Solve - ~ x + 4 x-3 g) Use the substitution

& Conducted I - CH - 11... · 6 01. y = x . tan x Differentiating wrt x ; SUMS ON PRODUCT RULE dy = x d tan x + tan x . d x dx dx dx

& Conducted I - CH - 11... · 6 01. y = x . tan x Differentiating wrt x ; SUMS ON PRODUCT RULE dy = x d tan x + tan x . d x dx dx dx

C4 Edexcel Solution Bank - Chapter 4...4 dx dt dy dt dy dx 4 sint cost dx dt dy dt dy dx sec2 t sec t tant sec t tant 1 cost cost sint 1 sint dx dt dy dt dy dx 2 sin2t 2 − 2 cos2t

C4 Edexcel Solution Bank - Chapter 4...4 dx dt dy dt dy dx 4 sint cost dx dt dy dt dy dx sec2 t sec t tant sec t tant 1 cost cost sint 1 sint dx dt dy dt dy dx 2 sin2t 2 − 2 cos2t

Mathematics Quadratic equationnflaw/EIE2106Sem12019-20/Tutorial1-Maths.pdf · Find the maximum/minimum Step 1: Find Step 2: Set 49 dy dx dy d xx x x x32 25101003 1010 dx dx 0 dy dx

Mathematics Quadratic equationnflaw/EIE2106Sem12019-20/Tutorial1-Maths.pdf · Find the maximum/minimum Step 1: Find Step 2: Set 49 dy dx dy d xx x x x32 25101003 1010 dx dx 0 dy dx

Drill: Find dy / dx

Drill: Find dy / dx

Xe .ie :±÷÷÷¥÷÷÷÷¥÷ii¥÷...dy y xy dx x xy y = 7) sin sin tan tan cos cos dy x y xy dx x y == 8) 1cos sin y y dy y e x dx xe x + = 9) 2 2 x y x y dy y y e dx yxe = 10)

Xe .ie :±÷÷÷¥÷÷÷÷¥÷ii¥÷...dy y xy dx x xy y = 7) sin sin tan tan cos cos dy x y xy dx x y == 8) 1cos sin y y dy y e x dx xe x + = 9) 2 2 x y x y dy y y e dx yxe = 10)

Calculus Differentiation dy/dx = y y = e x e ix = cosx + i sinx.

Calculus Differentiation dy/dx = y y = e x e ix = cosx + i sinx.

Differential Equationsdx dy + 2 =-ans: y=e-x+ce-2x 2. 2 sin 3 x x y dx dy x + = ans: x3y=c-cosx 3. xdy + ydx = ydy ans: y y c x = + 2 4- Exact The equation M(x, y)dx + N(x, y)dy =

Differential Equationsdx dy + 2 =-ans: y=e-x+ce-2x 2. 2 sin 3 x x y dx dy x + = ans: x3y=c-cosx 3. xdy + ydx = ydy ans: y y c x = + 2 4- Exact The equation M(x, y)dx + N(x, y)dy =

TRANSFORMS AND PARTIAL DIFFERENTIAL …...11. Find the general solution of px+qy=z Answer: The auxiliary equation is dx dy dz x y z From dx dy xy Integrating we get log x = log y +

TRANSFORMS AND PARTIAL DIFFERENTIAL …...11. Find the general solution of px+qy=z Answer: The auxiliary equation is dx dy dz x y z From dx dy xy Integrating we get log x = log y +

Lesson 8 - justcalculus · 2016. 12. 23. · Summary: The Flow of A Vector Field ALONG a Closed Curve: R C ³ ³³m(x,y)dx n(x,y)dy rotField(x,y)dx dy Let C be a closed curve parameterized

Lesson 8 - justcalculus · 2016. 12. 23. · Summary: The Flow of A Vector Field ALONG a Closed Curve: R C ³ ³³m(x,y)dx n(x,y)dy rotField(x,y)dx dy Let C be a closed curve parameterized

4 !: :wu ü - GitHub Pages · « 4. 1 D = {(x ,y) | 1 ! x ! 2, 0 ! y ! 1} qb qz! ! D (x 2 + x y + y2)dx dy =! 2 1" ! 1 0 (x 2 + x y + y2)dy dx =! 2 1 (x 2 y + 1 2 x y2 + 1 3 y3) 1

4 !: :wu ü - GitHub Pages · « 4. 1 D = {(x ,y) | 1 ! x ! 2, 0 ! y ! 1} qb qz! ! D (x 2 + x y + y2)dx dy =! 2 1" ! 1 0 (x 2 + x y + y2)dy dx =! 2 1 (x 2 y + 1 2 x y2 + 1 3 y3) 1

Potential Energy Length hence dl-dx = (1/2) (dy/dx) 2 dx dU = (1/2) F (dy/dx) 2 dx potential energy of element dx y(x,t)= y m sin( kx- t) dy/dx= y m.

Potential Energy Length hence dl-dx = (1/2) (dy/dx) 2 dx dU = (1/2) F (dy/dx) 2 dx potential energy of element dx y(x,t)= y m sin( kx- t) dy/dx= y m.

HIGHLIGHTS - jkshahclasses.com - … · 8 10. y = (9x3 1). tan x Differentiating wrt x ; dy = (9x3 1). d tan x + tan x d (9x3 1) dx dx dx = (9x3 1) . sec2 x + tan x . 27x2 = (9x3

HIGHLIGHTS - jkshahclasses.com - … · 8 10. y = (9x3 1). tan x Differentiating wrt x ; dy = (9x3 1). d tan x + tan x d (9x3 1) dx dx dx = (9x3 1) . sec2 x + tan x . 27x2 = (9x3

Answers - Nulake · 36 99 36 13 Page 38 186. dy dx = 2x cos x – x2 sin x 187. dy dx = 3 tan 3x + (9x – 6) sec2 3x 188. g’(x) = dy dx = 2x cot(5x – 1)

Answers - Nulake · 36 99 36 13 Page 38 186. dy dx = 2x cos x – x2 sin x 187. dy dx = 3 tan 3x + (9x – 6) sec2 3x 188. g’(x) = dy dx = 2x cot(5x – 1)