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Formulas in Pure Math 3

Formulas in Pure Math 3

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Mathematics, Other

12th Grade

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KASSIA! LLTTF

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1

Formulas in Pure Math 3

Module 3

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2

Table of contents

Limits : Slide 3

Differentiation : Slide 7

Integration : Slide 11

3

 Limits 

  • (a) If k is a constant ,  limxak = k\lim_{x\rightarrow a}k\ =\ k  

  • (b)  limxak(f(x)) = k ×limxaf(x) \lim_{x\rightarrow a}k\left(f\left(x\right)\right)\ =\ k\ \times\lim_{x\rightarrow a}f\left(x\right)\   

  • (c)  limxa[f(x)+g(x)]=limxaf(x)+limxag(x)\lim_{x\rightarrow a}\left[f\left(x\right)+g\left(x\right)\right]=\lim_{x\rightarrow a}f\left(x\right)+\lim_{x\rightarrow a}g\left(x\right)  

  •  (d) limxa[f(x)g(x)]=limxaf(x) limxag(x)\left(d\right)\ \lim_{x\rightarrow a}\left[f\left(x\right)-g\left(x\right)\right]=\lim_{x\rightarrow a}f\left(x\right)\ -\lim_{x\rightarrow a}g\left(x\right)  

  • (e)  limxa[f(x) ×g(x)]=limxaf(x) ×limxag(x)\lim_{x\rightarrow a}\left[f\left(x\right)\ \times g\left(x\right)\right]=\lim_{x\rightarrow a}f\left(x\right)\ \times\lim_{x\rightarrow a}g\left(x\right)  

  • (f)  limxa f(x)g(x)=limxaf(x)limxag(x) \lim_{x\rightarrow a}\ \frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\rightarrow a}f\left(x\right)}{\lim_{x\rightarrow a}g\left(x\right)}\   

  • (g)  limxa[f(x)]n=[limxaf(x)]n\lim_{x\rightarrow a}\left[f\left(x\right)\right]^n=\left[\lim_{x\rightarrow a}f\left(x\right)\right]^n  (assuming that the nth power is defined.

4

continued

  •  limxa nf(x)=nlimxaf(x)\lim_{x\rightarrow a}\ ^n\sqrt{f\left(x\right)}=^n\sqrt{\lim_{x\rightarrow a}f\left(x\right)}  (Assuming that ythe nth root is defined. )

  •  Note :Note\ :  

  •  limxax =a\lim_{x\rightarrow a}x\ =a  

  •  limx0 sinxx=1\lim_{x\rightarrow0}\ \frac{\sin x}{x}=1   ( limx0sinaxax\lim_{x\rightarrow0}\frac{\sin ax}{ax}   Also counts )

  •  limx0 1cosxx=0\lim_{x\rightarrow0}\ \frac{1-\cos x}{x}=0  

  •  limx+(1+ 1x)x=e\lim_{x\rightarrow+\infty}\left(1+\ \frac{1}{x}\right)^x=e  

5

 

L'Hopital's  Rule 

Limits of the form   lim f(x)g(x)\lim\ \frac{f\left(x\right)}{g\left(x\right)}   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 f(x)g(x)=lim f1(x)g1(x),\lim\ \frac{f\left(x\right)}{g\left(x\right)}=\lim\ \frac{f^1\left(x\right)}{g^1\left(x\right)},  
where   f1(x)f^1\left(x\right)   and  g1(x)g^1\left(x\right)   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.  limxa f(x) = L must exist\lim_{x\rightarrow a}\ f\left(x\right)\ =\ L\ must\ exist  
2. f (a) must exist 
3. f (a) = L 

 

7

 Differentiation 

 9. Differentiation  y=xn dydx=nxn19.\ Differentiation\ \cdot\ y=x^n\rightarrow\ \frac{dy}{dx}=nx^{n-1}  
  y=axn  dydx=anxn1\cdot\ y=ax^n\ \rightarrow\ \frac{dy}{dx}=anx^{n-1}  
  y=+ or  (whole number)x  dydx=whole number\cdot\ y=+\ or\ -\ \left(whole\ number\right)x\ \rightarrow\ \frac{dy}{dx}=whole\ number    y=+ or  whole number   dydx=0\cdot\ y=+\ or\ -\ whole\ number\ \ \rightarrow\ \frac{dy}{dx}=0  
  y=sinx  dydx=cos x\cdot\ y=\sin x\ \rightarrow\ \frac{dy}{dx}=\cos\ x  
 y=cos x  dydx= sinx y=\cos\ x\ \rightarrow\ \frac{dy}{dx}=-\ \sin x\   

 y=sin (ax+b) dydx=acos (ax+b)\cdot y=\sin\ \left(ax+b\right)\rightarrow\ \frac{dy}{dx}=a\cos\ \left(ax+b\right)    y=cos (ax+b)=sin (ax+b)×a\cdot\ y=\cos\ \left(ax+b\right)=-\sin\ \left(ax+b\right)\times a   y=cos (f(x)) dydx=sin(f(x))×f1(x)\cdot y=\cos\ \left(f\left(x\right)\right)\rightarrow\ \frac{dy}{dx}=-\sin\left(f\left(x\right)\right)\times f^1\left(x\right)    y=sin (f(x)) dydx=cos (f(x)) ×f1(x)\cdot\ y=\sin\ \left(f\left(x\right)\right)\rightarrow\ \frac{dy}{dx}=\cos\ \left(f\left(x\right)\right)\ \times f^1\left(x\right)  

8

*   (ax+b)nn(ax+b)n1×a\left(ax+b\right)^n\rightarrow n\left(ax+b\right)^{n-1}\times a  

 tanxsec2x\tan x\rightarrow\sec^2x  
 tan[f(x)]=sec2[f(x)]×f(x)\tan\left[f\left(x\right)\right]=\sec^2\left[f\left(x\right)\right]\times f'\left(x\right)  
 tan(ax+b)sec2(ax+b)×a\tan\left(ax+b\right)\rightarrow\sec^2\left(ax+b\right)\times a  
 af(x)af(x)a\cdot f\left(x\right)\rightarrow a\cdot f'\left(x\right) 

9

 

  • Product Rule :  dydx=U dvdx+V dudx\frac{dy}{dx}=U\ \frac{dv}{dx}+V\ \frac{du}{dx}   

  • Quotient Rule :  dydx=V dudxU dvdxV2\frac{dy}{dx}=\frac{V\ \frac{du}{dx}-U\ \frac{dv}{dx}}{V^2}  

  • Chain Rule :  y=f (g(x))  dydx=f1(g(x)) g1(x)y=f\ \left(g\left(x\right)\right)\ \rightarrow\ \frac{dy}{dx}=f^{-1}\left(g\left(x\right)\right)\ g^{-1}\left(x\right)  where  f1& g1f^{-1}\&\ g^{-1}  are the differentiated functions / the derivatives.

  •  (ax+ b)n=n(ax+b)n1(a)\left(ax+\ b\right)^n=n\left(ax+b\right)^{n-1}\left(a\right)  

  •  At a S.p dydx=0   d2ydx2:2nd derivative testAt\ a\ S.p\ \frac{dy}{dx}=0\ \ \ \frac{d^2y}{dx^2}:2nd\ derivative\ test  (Maximum, Minimum, Point of inflection)    

  •  Increasing function  dydx>0     Decreasing function dydx<0\frac{dy}{dx}>0\ \ \ \ \ Decrea\sin g\ function\ \frac{dy}{dx}<0  

10

 x=f(t) and y = g(t)x=f\left(t\right)\ and\ y\ =\ g\left(t\right)  
 dydx=dydtdxdt=g(t)f(t)\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{g'\left(t\right)}{f'\left(t\right)}  

* First Principles

 limh0 f(x+h)f(x)h\lim_{h\rightarrow0}\ \frac{f\left(x+h\right)-f\left(x\right)}{h}  

11

Integration

  •  xndx = xn+1n+1+c\int_{ }^{ }x^ndx\ =\ \frac{x^{n+1}}{n+1}+c  

  •  axndx=axndx=a(xn+1n+1)+c\int_{ }^{ }ax^ndx=a\int_{ }^{ }x^ndx=a\left(\frac{x^{n+1}}{n+1}\right)+c  : where a is a constant and placed outside the integral sign to be multiplied by the integrated function later.

  •  (ax+b)n dx=(ax+b)n+1n+1×(1a)+c\int_{ }^{ }\left(ax+b\right)^n\ dx=\frac{\left(ax+b\right)^{n+1}}{n+1}\times\left(\frac{1}{a}\right)+c  

  •  cos x dx =sinx +c\int_{ }^{ }\cos\ x\ dx\ =\sin x\ +c  

  •   aax+c\int_{ }^{ }a\rightarrow ax+c  

12

continued..

  •  sinx dx=cos x+c\int_{ }^{ }-\sin x\ dx=\cos\ x+c  

  •  sinx dx =cosx+c\int_{ }^{ }\sin x\ dx\ =-\cos x+c  

  •  sin (ax+b)dx=cos(ax+b)a+c\int_{ }^{ }\sin\ \left(ax+b\right)dx=\frac{-\cos\left(ax+b\right)}{a}+c  

  •   cos(ax+b)dx=sin(ax+b)a+c\int\ \cos\left(ax+b\right)dx=\frac{\sin\left(ax+b\right)}{a}+c  

  •  sec2(x)tanx+c\int_{ }^{ }\sec^2\left(x\right)\rightarrow\tan x+c  

  •  sec2(ax+b) tan(ax+b)a+c\int_{ }^{ }\sec^2\left(ax+b\right)\rightarrow\ \frac{\tan\left(ax+b\right)}{a}+c  

13

 (an+b)n 1a (ax+b)n+1n+1+c , n1\int_{ }^{ }\left(an+b\right)^n\rightarrow\ \frac{1}{a}\ \frac{\left(ax+b\right)^{n+1}}{n+1}+c\ ,\ n\ne1  

 af(x)dxaf(x)dx\int_{ }^{ }a\cdot f\left(x\right)dx\equiv a\int_{ }^{ }f\left(x\right)dx  
 xy[f(x) +g(x)]dx=xyf(x)dx +g(x)dx\int_x^y\left[f\left(x\right)\ \frac{+}{ }g\left(x\right)\right]dx=\int_x^yf\left(x\right)dx\ \frac{+}{ }g\left(x\right)dx  
*  1xdx=lnx+c\int_{ }^{ }\ \frac{1}{x}dx=\ln x+c  
 cos ax  sinaxa+c\int_{ }^{ }\cos\ ax\ \rightarrow\ \frac{\sin ax}{a}+c  
* sin ax= cosaxa+c\int_{ }^{ }\sin\ ax=\ \frac{-\cos ax}{a}+c  
 To integrate sin2x change to  12(1cos2x)dx   [Identity used cos2x=12sin2x]To\ integrate\ \int_{ }^{ }\sin^2x\ change\ to\ \int_{ }^{ }\ \frac{1}{2}\left(1-\cos2x\right)dx\ \ \ \left[Identity\ used\ \cos2x=1-2\sin^2x\right]  
 To integrate cos2x dx change to  12(1+cos2x)dx [Identity used cos2x=2cos2x1]To\ integrate\ \int\cos^2x\ dx\ change\ to\ \int\ \frac{1}{2}\left(1+\cos2x\right)dx\ \left[Identity\ used\ \cos2x=2\cos^2x-1\right]  

14

* To integrate  tan2x dx change to (sec2x1)dx   [Identity used 1+tan2x=sec2x]\int\tan^2x\ dx\ change\ to\ \int\left(\sec^2x-1\right)dx\ \ \ \left[Identity\ used\ 1+\tan^2x=\sec^2x\right]  

* To integrate  sinxcos change to  (12sin 2x) dx   [Identity used sin 2x=2sin x cos x]\int\sin x\cos\ change\ to\ \int\ \left(\frac{1}{2}\sin\ 2x\right)\ dx\ \ \ \left[Identity\ used\ \sin\ 2x=2\sin\ x\ \cos\ x\right]  

15

Definite Integrals

 f(x)dx=g(x)ab =g(b)g(a)\int f\left(x\right)dx=\lceil g\left(x\right)\rceil_a^b\ =g\left(b\right)-g\left(a\right)  
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\int_a^bf\left(x\right)+\int_b^cf\left(x\right)\ dx=\int_a^cf\left(x\right)\ dx  

Area Under A curve 
 Region bounded by (above)the x axis : A=aby dxRegion\ bounded\ by\ \left(above\right)the\ x\ axis\ :\ A=\int_a^by\ dx    Region is under the x axis: A =x1x2y dx Region\ is\ under\ the\ x\ axis:\ A\ =-\int_{x_1}^{x_2}y\ dx\      Region is on the left side of the y axis (bounded by the y axis not x axis) : A=y1y2xdy\ Region\ is\ on\ the\ left\ side\ of\ the\ y\ axis\ \left(bounded\ by\ the\ y\ axis\ not\ x\ axis\right)\ :\ A=-\int_{y_1}^{y_2}xdy   Region on the right of the y axis (bounded by the y axis ): A =y1y2 x dyRegion\ on\ the\ right\ of\ the\ y\ axis\ \left(bounded\ by\ the\ y\ axis\ \right):\ A\ =\int_{y_1}^{y_2}\ x\ dy  
Same process as definite integrals except the numerical answer is  2 units ^2\ units\   because the "area" of the region was found. 

 

16

* Region rotated about the x axis :  V=πx1x2y2dxV=\pi\int_{x_1}^{x_2}y^2dx  
*Region rotated about the y axis :  V=πy1y2x2dyV=\pi\int_{y_1}^{y_2}x^2dy  

 xaf(x)d=xaf(ax)dx, a>0\int_x^af\left(x\right)d=\int_x^af\left(a-x\right)dx,\ a>0  

Formulas in Pure Math 3

Module 3

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