MATH 221

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

Last updated: 4 August 2014

Lecture 10

Finding derivatives with limits

If f is a function then f(x)=f(a)+ (dfdx|x=a) (x-a)+12! (d2fdx2|x=a) (x-a)2+13! (d3fdx3|x=a) (x-a)3+. Substitute x=a+Δx. Δx stands for a small change in x. f(a+Δx)= f(a)+(dfdx|x=a) (Δx)+12! (d2fdx2|x=a) (Δx)2+13! (d3fdx3|x=a) (Δx)3+. So f(a+Δx)-f(a) =(dfdx|x=a) Δx+12! (d2fdx2|x=a) (Δx)2+. So f(a+Δx)-f(a)Δx =dfdx|x=a+ 12! (d2ddx2|x=a) Δx+13! (d3fdx3|x=a) (Δx)2+. So limΔx0 f(a+Δx)-f(a) Δx =dfdx|x=a.

Suppose f(x)=x3. What is f(3.02)? f(3.02) = f(3+.02) = f(3)+ (dfdx) (.02)+12 (d2fdx2) (.02)2+ since f(a+Δx)=f (a)+(dfdx|x=a) Δx+12(d2fdx2|x=a) (Δx)2+. Now f(3)=27, dfdx|x=3 =3x2|x=3=27, d2fdx2 |x=3=6x|x=3 =18, d3fdx3 |x=3=6|x=3 =6, d4fdx4|x=3 =0|x=3=0, d5fdx5|x=3 =0|x=3=0, So f(3.02) = f(3+.02) = 27+27(.02)+12 18(.02)2+63! (.02)3+0+0+ = 27+.54+9(.0004)+ .000008 = 27.543608.

What is the expansion of f(a+Δx) when f(x)=e3x and a=0? f(a+Δx) = e3(a+Δx) =e3Δx = 1+3Δx+ (3Δx)22! +(3Δx)33! +(3Δx)44!+ = 1+3Δx+92! (Δx)2+273! (Δx)3+ 34(Δx)44!+. Second way: f(a+Δx) = f(0+Δx) = f(0)+dfdx |x=0Δx+ 12(d2fdx2|x=0) (Δx)2+. f(0)=e3·0= 1, dfdx|x=0 =de3xdx |x=0=3e3x |x=0=3, d2fdx2 |x=0=d3e3xdx |x=0=32e3x |x=0=32, d3fdx3 |x=0= d32e3xdx |x=0=33 e3x|x=0= 33,. So f(0+Δx)= e3Δx=1+3 Δx+1232 (Δx)2+ 13!33 (Δx)3+.

If f(x)=ln(1+x) expand f(a+Δx) in terms of Δx when a=0. f(a+Δx) = f(0+Δx)= f(Δx)=ln(1+Δx) = f(0)+ (dfdx|x=0) Δx+12 (d2fdx2|x=0) (Δx)2+13! (d3fdx3|x=0) (Δx)3+ = ln(1+0)= dln(1+x)dx |x=0+12! (d2ln(1+x)dx2|x=0) ()2+, ln(1+0) = ln1=0, dln(1+x)dx |x=0 = 11+x|x=0 =11=1, d2ln(1+x)dx2 |x=0 = d11+xdx |x=0= -1(1+x)2 |x=0=-1, d3ln(1+x)dx3 |x=0 = d-1(1+x)2dx |x=0= (-1)(-2) (1+x)3 |x=0=2·1, d4ln(1+x)dx4 |x=0 = d2!(1+x)3dx |x=0= -3·2·1(1+x)4 |x=0=-3!. So ln(1+Δx) = 0+(+1)Δx+ 12!(-1) (Δx)2+ 13!2! (Δx)3+ 14!(-3!) (Δx)4+ = 0+Δx- (Δx)22+ (Δx)33- (Δx)44+ = Δx- (Δx)22+ (Δx)33- (Δx)44+. The linear approximation to ln(1+x) at x=0 is ln(1+Δx) +Δx. The quadratic approximation to ln(1+x) at x=0 is ln(1+Δx) Δx-(Δx)22!.

Approximate the value of (255)14.

Let f(x)=x14. Then (255)14= (256-1)14= f(a+Δx) with a=256 and Δx=-1. f(a+Δx) f(a)+ (dfdx|x=a) Δx=f(256)+ (dfdx|x=256) (-1) = (256)14+ (14x-34|x=256) (-1) = 4+14(256)-34 (-1)=4+14 4-3(-1) = 4+14143(-1) =4-1256=3255256= 3.99609375. This is an approximation to 25514 using a linear approximation. f(a+Δx) f(a)+ (dfdx|x=a) Δx+12 (d2fdx2|x=a) (Δx)2 = a14+ (14x-34|x=a) Δx+12(14) (-34)x-74 |x=a(Δx)2 = a14+14 (a14)-3Δx +-32142 (a14)-7 (Δx)2 = 4+144-3(-1) -32·424-7 (-1)2 = 4-144+ 32·49= 4-116+ 32.262144 = 3.99609947204589843750. This is the quadratic approximation to 25514. The correct answer is 25514=3.9960880148804670689 according to my computer. The computer is clearly wrong (it is off by at least .000005).

Expand f(a+Δx) in terms of Δx when a=4 and f(x)=x. f(a+Δx) = f(4+Δx)= 4+Δx = f(4)+ (dfdx|x=4) Δx+12! (d2fdx2|x=4) (Δx)2+, f(4) = 4=2. dfdx |x=4 = dx12dx |x=4=12 x-12|x=4 =12·12=14, d2fdx2 |x=4 = d12x-12dx |x=4=12 (-12)x-32 |x=4=-12·2 123=-125, d3fdx3 |x=4 = d-122x-32dx |x=4=-122 -32x-52 |x=4=323 125=328, d4fdx4 |x=4 = d323x-52dx |x=4=323 (-52)x-72 |x=4=-3·524 127=-3·5211, So 4+Δx = f(4+Δx) = 2+122Δx -12!125 (Δx)2+13! 328(Δx)3 -14!3·5211 (Δx)4+. The linear approximation to x at x=4 is 4+Δx2+ 14Δx. The quadratic approximation to x at x=4 is 4+Δx2+ 14Δx-164 (Δx)2.

Approximate 4.03.

Linear: 4.032+14 (.03)=2.0075. Quadratic: 4.032+14(.03) -164(.03)2= 2.0074859375.

Notes and References

These are a typed copy of Lecture 10 from a series of handwritten lecture notes for the class MATH 221 given on September 29, 2000.

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