MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 28 July 2014

Lecture 3

A function takes in a number, chews on it and spits out another number input numberx f output numberf(x). A constant function always spits out the same number, no matter what the input is.

f(x)=2. x f 2 If x=3π+7 then f(x)=2.
If x=3+3i then f(x)=2.
If x=1 then f(x)=2.
If x=7 then f(x)=2.

We call this function 2. 5 2 2 10 2 2 So 2 sometimes means the number 2 and sometimes means the function 2.

Derivatives

A derivative takes in a function, chews on it and spits out another function f ddx dfdx. The derivative ddx spits out a function according to the rules:

(1) dxdx=1.
(2) d(cf)dx=cdfdx, if c does not change when x changes.
(3) d(f+g)dx=dfdx+dgdx.
(4) d(fg)dx=fdgdx+dfdxg.

Find dydx if y=5x. dydx= d(5x)dx= 5dxdx=5·1 =5.

Find dydx if y=πx. dydx= d(πx)dx= πdxdx= π·1=π.

Find dydx if y=1. dydx= d1dx= d(1·1)dx= 1·d1dx+d1dx·1= d1dx+d1dx. Subtract d1dx from both sides. So d1dx=0.

Find dydx if y=5. dydx= d5dx= d(5·1)dx= 5d1dx= 5·0=0.

Find dydx if y=6342. d6342dx= d(6342·1)dx= 6342d1dx=6342·0=0.

Find dcdx if c is a constant. dcdx= d(c·1)dx= cd1dx=c·0=0.

Find dydx if y=3x+12. dydx= d(3x+12)dx= d(3x)dx+ d12dx= 3dxdx+0= 3·1+0=3.

Find dydx if y=x2. dydx= dx2dx= dx·xdx= x·dxdx+= x·1+1·x=2x. dxdx·x

Find dx3dx. dx3dx= d(x2·x)dx= x2dxdx+ dx2dx·x= x2·1+2x·x=3 x2.

Find dx4dx. dx4dx= d(x3·x)dx= x3·dxdx+ dx3dx·x= x3·1+3x2·x= 4x3.

Find dx6342dx. dx6342dx= d(x6341·x)dx= x6341·dxdx+ dx6341dx·x= x6341·1+6341x6340·x= 6342x6341.

Find dxndx if n=1,2,3,. dxndx = d(xn-1·x)dx = xn-1·dxdx+ dnn-1dx·x = xn-1·1+ (n-1)xn-2·x ( since we just found dxn-1dx= (n-1)xn-2 ) = xn-1+(n-1) xn-1. So dxndx= nxn-1.

Find dxndx if n=0. dxndx= dx0dx= d1dx=0=n xn-1.

Find dx-6342dx. d(x-6342·x6342)dx= dx0dx=d1dx =0. On the other hand, d(x-6342·x6342)dx = x-6342 dx6342dx+ dx-6342dx ·x6342 = x-63426342x6341+ d-6342dx· x6342. So 0=6342x-1+x6342 dx-6342dx. So dx-6342dx = -6342x-1 x-6342 = (-6342) x-6342.

Find dx-ndx if n=1,2,3,. dxnx-ndx = xndx-ndx+ dxndxx-n = xndx-ndx +nxn-1x-n = xndx-ndx +nx-1. On the other hand d(xnx-n)dx =dx0dx= d1dx=0. So xndx-ndx +nz-1=0. Solve for dx-ndx. Then dx-ndx= -nx-1x-n= -nx-n-1.

Let y=3x3+5x2+2x+7. Find dydx. dydx = d(3x3+5x2+2x+7>)dx = d(3x3)dx+ d(5x2+2x+7)dx = d(3x3)dx+ d(5x2)dx+ d(2x)dx+ d7dx = 3dx3dx+ 5dx2dx+ 2dxdx+ 7d1dx = 3·3x2+ 5·2x+ 2·1+ 7·0 = 9x2+10x+2.

If y=-7x-13+5x-7+(6+2i)x38. Find dydx. dydx = d(-7x-13+5x-7+(6+2i)x38)dx = d(-7x-13)dx+ d(5x-7)dx+ d((6+2i)x38)dx = -7dx-13dx+ 5dx-7dx+ (6+2i)dx38dx = (-7)(-13)x-13-1+ 5(-7)x-7-1+ (6+2i)38x38-1 = 91x-14-35x-8 +(228+76i)x37.

Notes and References

These are a typed copy of Lecture 3 from a series of handwritten lecture notes for the class MATH 221 given on September 11, 2000.

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