

Formulas in Pure Math 3
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Mathematics, Other
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12th Grade
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Hard
KASSIA! LLTTF
Used 29+ times
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16 Slides • 0 Questions
1
Formulas in Pure Math 3
Module 3

2
Table of contents
Limits : Slide 3
Differentiation : Slide 7
Integration : Slide 11
3
Limits
(a) If k is a constant , x→alimk = k
(b) x→alimk(f(x)) = k ×x→alimf(x)
(c) x→alim[f(x)+g(x)]=x→alimf(x)+x→alimg(x)
(d) x→alim[f(x)−g(x)]=x→alimf(x) −x→alimg(x)
(e) x→alim[f(x) ×g(x)]=x→alimf(x) ×x→alimg(x)
(f) x→alim g(x)f(x)=limx→ag(x)limx→af(x)
(g) x→alim[f(x)]n=[x→alimf(x)]n (assuming that the nth power is defined.
4
continued
x→alim nf(x)=nx→alimf(x) (Assuming that ythe nth root is defined. )
Note :
x→alimx =a
x→0lim xsinx=1 ( x→0limaxsinax Also counts )
x→0lim x1−cosx=0
x→+∞lim(1+ x1)x=e
5
L'Hopital's Rule
Limits of the form lim g(x)f(x) can be evaluated by this rule (if they exist) in the indeterminate cases where f(x) and g(x) both approach 0 or both approach +infinity or - infinity . The rule states :
lim g(x)f(x)=lim g1(x)f1(x),where f1(x) and g1(x) are the first derivatives of f(x) and g(x) respectively, with respect to x.
6
Continuity
A function f(x) is said to be continuous at x = a if
1. x→alim f(x) = L must exist
2. f (a) must exist
3. f (a) = L
7
Differentiation
9. Differentiation ⋅ y=xn→ dxdy=nxn−1
⋅ y=axn → dxdy=anxn−1
⋅ y=+ or − (whole number)x → dxdy=whole number ⋅ y=+ or − whole number → dxdy=0
⋅ y=sinx → dxdy=cos x
y=cos x → dxdy=− sinx
8
* (ax+b)n→n(ax+b)n−1×a
* tanx→sec2x* tan[f(x)]=sec2[f(x)]×f′(x)
* tan(ax+b)→sec2(ax+b)×a
* a⋅f(x)→a⋅f′(x)
9
Product Rule : dxdy=U dxdv+V dxdu
Quotient Rule : dxdy=V2V dxdu−U dxdv
Chain Rule : y=f (g(x)) → dxdy=f−1(g(x)) g−1(x) where f−1& g−1 are the differentiated functions / the derivatives.
(ax+ b)n=n(ax+b)n−1(a)
At a S.p dxdy=0 dx2d2y:2nd derivative test (Maximum, Minimum, Point of inflection)
Increasing function dxdy>0 Decreasing function dxdy<0
10
dxdy=dtdxdtdy=f′(t)g′(t)
* First Principles
11
Integration
∫xndx = n+1xn+1+c
∫axndx=a∫xndx=a(n+1xn+1)+c : where a is a constant and placed outside the integral sign to be multiplied by the integrated function later.
∫(ax+b)n dx=n+1(ax+b)n+1×(a1)+c
∫cos x dx =sinx +c
∫a→ax+c
12
continued..
∫−sinx dx=cos x+c
∫sinx dx =−cosx+c
∫sin (ax+b)dx=a−cos(ax+b)+c
∫ cos(ax+b)dx=asin(ax+b)+c
∫sec2(x)→tanx+c
∫sec2(ax+b)→ atan(ax+b)+c
13
* ∫(an+b)n→ a1 n+1(ax+b)n+1+c , n=1
* ∫a⋅f(x)dx≡a∫f(x)dx* ∫xy[f(x) +g(x)]dx=∫xyf(x)dx +g(x)dx
* ∫ x1dx=lnx+c
* ∫cos ax → asinax+c
* ∫sin ax= a−cosax+c
* To integrate ∫sin2x change to ∫ 21(1−cos2x)dx [Identity used cos2x=1−2sin2x]
* To integrate ∫cos2x dx change to ∫ 21(1+cos2x)dx [Identity used cos2x=2cos2x−1]
14
* To integrate ∫tan2x dx change to ∫(sec2x−1)dx [Identity used 1+tan2x=sec2x]
* To integrate ∫sinxcos change to ∫ (21sin 2x) dx [Identity used sin 2x=2sin x cos x]15
Definite Integrals
where f (x) dx is the derivative to be integrated , g (x) is the integrated f (x) & a and b are the limits.
* If a < b < c, ∫abf(x)+∫bcf(x) dx=∫acf(x) dx
Area Under A curve
Region bounded by (above)the x axis : A=∫aby dx Region is under the x axis: A =−∫x1x2y dx Region is on the left side of the y axis (bounded by the y axis not x axis) : A=−∫y1y2xdy Region on the right of the y axis (bounded by the y axis ): A =∫y1y2 x dy
Same process as definite integrals except the numerical answer is 2 units because the "area" of the region was found.
16
* Region rotated about the x axis : V=π∫x1x2y2dx
*Region rotated about the y axis : V=π∫y1y2x2dy
Formulas in Pure Math 3
Module 3

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