12X1 T01 02 differentiating logs

33 %
67 %
Information about 12X1 T01 02 differentiating logs
Education

Published on November 19, 2009

Author: nsimmons

Source: slideshare.net

Differentiating Logarithms

Differentiating Logarithms y  log f  x 

Differentiating Logarithms y  log f  x  dy f  x   dx f  x 

Differentiating Logarithms y  log f  x  y  log a f  x  dy f  x   dx f  x 

Differentiating Logarithms y  log f  x  y  log a f  x  dy f  x  dy f  x    dx f  x  dx log a  f  x 

Differentiating Logarithms y  log f  x  y  log a f  x  dy f  x  dy f  x    dx f  x  dx log a  f  x  e.g. (i) y  log3 x  5

Differentiating Logarithms y  log f  x  y  log a f  x  dy f  x  dy f  x    dx f  x  dx log a  f  x  e.g. (i) y  log3 x  5 dy 3  dx 3 x  5

Differentiating Logarithms y  log f  x  y  log a f  x  dy f  x  dy f  x    dx f  x  dx log a  f  x  e.g. (i) y  log3 x  5 ii  y  log x 3 dy 3  dx 3 x  5

Differentiating Logarithms y  log f  x  y  log a f  x  dy f  x  dy f  x    dx f  x  dx log a  f  x  e.g. (i) y  log3 x  5 ii  y  log x 3 dy 3 dy 3 x 2   3 dx 3 x  5 dx x

Differentiating Logarithms y  log f  x  y  log a f  x  dy f  x  dy f  x    dx f  x  dx log a  f  x  e.g. (i) y  log3 x  5 ii  y  log x 3 dy 3 dy 3 x 2   3 dx 3 x  5 dx x 3  x

Differentiating Logarithms y  log f  x  y  log a f  x  dy f  x  dy f  x    dx f  x  dx log a  f  x  e.g. (i) y  log3 x  5 ii  y  log x 3 OR y  log x 3 2 dy 3 dy 3 x   3 dx 3 x  5 dx x 3  x

Differentiating Logarithms y  log f  x  y  log a f  x  dy f  x  dy f  x    dx f  x  dx log a  f  x  e.g. (i) y  log3 x  5 ii  y  log x 3 OR y  log x 3 2 dy  3 dy 3 x  3 y  3 log x dx 3 x  5 dx x 3  x

Differentiating Logarithms y  log f  x  y  log a f  x  dy f  x  dy f  x    dx f  x  dx log a  f  x  e.g. (i) y  log3 x  5 ii  y  log x 3 OR y  log x 3 2 dy  3 dy 3 x  3 y  3 log x dx 3 x  5 dx x dy 3 3   dx x x

(iii) y  loglog x 

(iii) y  loglog x  1 dy  x dx log x

(iii) y  loglog x  1 dy  x dx log x 1  x log x

(iii) y  loglog x  1 dy  x dx log x 1  x log x iv  y  log x  3 x  2

(iii) y  loglog x  1 dy  x dx log x 1  x log x iv  y  log x  3 x  2 dy  x  31   x  2 1  dx  x  3 x  2

(iii) y  loglog x  1 dy  x dx log x 1  x log x iv  y  log x  3 x  2 dy  x  31   x  2 1  dx  x  3 x  2 2x  5   x  3 x  2

(iii) y  loglog x  1 dy  x dx log x 1  x log x iv  y  log x  3 x  2 OR y  log x  3  log x  2  dy  x  31   x  2 1  dx  x  3 x  2 2x  5   x  3 x  2

(iii) y  loglog x  1 dy  x dx log x 1  x log x iv  y  log x  3 x  2 OR y  log x  3  log x  2  dy  x  31   x  2 1 dy 1 1    dx  x  3 x  2 dx x  3 x  2 2x  5   x  3 x  2

(iii) y  loglog x  1 dy  x dx log x 1  x log x iv  y  log x  3 x  2 OR y  log x  3  log x  2  dy  x  31   x  2 1 dy 1 1    dx  x  3 x  2 dx x  3 x  2 2x  5  x  2   x  3    x  3 x  2  x  3 x  2

(iii) y  loglog x  1 dy  x dx log x 1  x log x iv  y  log x  3 x  2 OR y  log x  3  log x  2  dy  x  31   x  2 1 dy 1 1    dx  x  3 x  2 dx x  3 x  2 2x  5  x  2   x  3    x  3 x  2  x  3 x  2 2x  5   x  3 x  2

v   x  5 y  log    x  2

v  y  log  x  5   x  2  x  21   x  51 dy   x  22 dx x5 x2

v  y  log  x  5   x  2  x  21   x  51 dy   x  22 dx x5 x2 3  x  2    x  22  x  5

v  y  log  x  5   x  2  x  21   x  51 dy   x  22 dx x5 x2 3  x  2    x  22  x  5 3   x  2 x  5

v  y  log  x  5  OR y  log x  5  log x  2   x  2  x  21   x  51 dy   x  22 dx x5 x2 3  x  2    x  22  x  5 3   x  2 x  5

v  y  log  x  5  OR y  log x  5  log x  2   x  2  x  21   x  51 dy  1  1 dy  x  22 dx x  5 x  2  dx x5 x2 3  x  2    x  22  x  5 3   x  2 x  5

v  y  log  x  5  OR y  log x  5  log x  2   x  2  x  21   x  51 dy  1  1 dy  x  22 dx x  5 x  2  dx x5  x  2    x  5  x2  x  5 x  2 3  x  2    x  22  x  5 3   x  2 x  5

v  y  log  x  5  OR y  log x  5  log x  2   x  2  x  21   x  51 dy  1  1 dy  x  22 dx x  5 x  2  dx x5  x  2    x  5  x2  x  5 x  2 3  x  2 3     x  22  x  5  x  2 x  5 3   x  2 x  5

v  y  log  x  5  OR y  log x  5  log x  2   x  2  x  21   x  51 dy  1  1 dy  x  22 dx x  5 x  2  dx x5  x  2    x  5  x2  x  5 x  2 3  x  2 3     x  22  x  5  x  2 x  5 3   x  2 x  5 vi  y  log10 6 x

v  y  log  x  5  OR y  log x  5  log x  2   x  2  x  21   x  51 dy  1  1 dy  x  22 dx x  5 x  2  dx x5  x  2    x  5  x2  x  5 x  2 3  x  2 3     x  22  x  5  x  2 x  5 3   x  2 x  5 vi  y  log10 6 x dy 6  dx log 10 6 x

v  y  log  x  5  OR y  log x  5  log x  2   x  2  x  21   x  51 dy  1  1 dy  x  22 dx x  5 x  2  dx x5  x  2    x  5  x2  x  5 x  2 3  x  2 3     x  22  x  5  x  2 x  5 3   x  2 x  5 vi  y  log10 6 x dy 6  dx log 10 6 x 1  x log 10

v  y  log  x  5  OR y  log x  5  log x  2   x  2  x  21   x  51 dy  1  1 dy  x  22 dx x  5 x  2  dx x5  x  2    x  5  x2  x  5 x  2 3  x  2 3     x  22  x  5  x  2 x  5 3   x  2 x  5 Exercise 12B; 1acf, 2chk, 5acehi, 6b, 7ad, 8acef, 9bd, 10ac, 11, 13a, 14bdfhjl, vi  y  log10 6 x 15b, 18bdf, 19b, 20af*, 21a* dy 6  dx log 10 6 x Exercise 12C; 1bdf, 2, 3, 6, 7a, 8, 11, 1  13, 14, 18* x log 10

Add a comment

Related presentations

Related pages

12X1 T01 02 differentiating logs - YouTube

Looking at the technique used when differentiating ... 12X1 T01 03 integrating ... virtualb15 34 views. 12:17 12X1 T04 02 growth ...
Read more

12X1 T01 logarithms - YouTube

12X1 T01 01 log laws ... 12X1 T01 02 differentiating logs by virtualb15. 6:49. Play next; Play now; 12X1 T01 03 integrating derivative on function
Read more

Yr 12 Mathematics - Mr Cameron - Google Sites

Mr Cameron. Year 11. Yr 11 Mathematics. Yr 11 Mathematics Ext 1. Yr 11 Maths General 2. Year 12. Yr 12 Mathematics. Yr 12 Mathematics Ext 1. Yr 12 ...
Read more

nsimmons

The technique of completing the square can be used when the equation cannot be factorised. As the steps are the same each time, this pattern lends itself ...
Read more

A Spec Wheels and Tires - SALE, 12x1.5 items in LUG NUTS ...

690 results found: Aodhan XT45 12X1.25 Racing Lug Nut Neo Chrome/Gol d Cap( Set Of 20Pc w/Key) · Aodhan XT45 12x1.25 Racing Lug Nut Chrome/Red Cap (Set Of ...
Read more

5. Derivative of the Logarithmic Function

... let's look at a graph of the log function with base e, that is: f ... Differentiating Logarithmic Functions with Bases other than ... 02 December 2014
Read more

24 Black Lug Nuts | eBay - Electronics, Cars, Fashion ...

Find great deals on eBay for 24 Black Lug Nuts in Wheel Lugs. Shop with confidence. Skip to main content. eBay: ... 12mm x 1.5 or 12x1.5 or M12x1.5.
Read more