Suppose a function F(z) is a product of \"n\" differentiable functions:
F(z)=f1(z)f2(z)f3(z)...fn(z).
Find F\'(z)/F(z) in simplest form.
(Its F prime in the numerator. Its not very clear because of the font.)
Solution
F(z)=f1(z)f2(z)f3(z)...fn(z) while differentiating one function will be differentiated
and others will remain constant F\'(z)=f1\'(z)f2(z)f3(z)...fn(z) + f1(z)f2\'(z)f3(z)...fn(z) +
f1(z)f2(z)f3\'(z)...fn(z) .................. + f1(z)f2(z)f3(z)...fn\'(z) F\'(z)/F(z) = [ f1\'(z)f2(z)f3(z)...fn(z)
+ f1(z)f2\'(z)f3(z)...fn(z) + f1(z)f2(z)f3\'(z)...fn(z) .................. +
f1(z)f2(z)f3(z)...fn\'(z)]/[f1(z)f2(z)f3(z)...fn(z)] F\'(z)/F(z) = f1\'(z)/f1(z) + f2\'(z)/f2(z) +
f3\'(z)/f3(z) + ...................... + fn\'(z)/fn(z).
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
Suppose a function F(z) is a product of n differentiable functio.pdf
1. Suppose a function F(z) is a product of "n" differentiable functions:
F(z)=f1(z)f2(z)f3(z)...fn(z).
Find F'(z)/F(z) in simplest form.
(Its F prime in the numerator. Its not very clear because of the font.)
Solution
F(z)=f1(z)f2(z)f3(z)...fn(z) while differentiating one function will be differentiated
and others will remain constant F'(z)=f1'(z)f2(z)f3(z)...fn(z) + f1(z)f2'(z)f3(z)...fn(z) +
f1(z)f2(z)f3'(z)...fn(z) .................. + f1(z)f2(z)f3(z)...fn'(z) F'(z)/F(z) = [ f1'(z)f2(z)f3(z)...fn(z)
+ f1(z)f2'(z)f3(z)...fn(z) + f1(z)f2(z)f3'(z)...fn(z) .................. +
f1(z)f2(z)f3(z)...fn'(z)]/[f1(z)f2(z)f3(z)...fn(z)] F'(z)/F(z) = f1'(z)/f1(z) + f2'(z)/f2(z) +
f3'(z)/f3(z) + ...................... + fn'(z)/fn(z)
![Suppose a function F(z) is a product of "n" differentiable functions:
F(z)=f1(z)f2(z)f3(z)...fn(z).
Find F'(z)/F(z) in simplest form.
(Its F prime in the numerator. Its not very clear because of the font.)
Solution
F(z)=f1(z)f2(z)f3(z)...fn(z) while differentiating one function will be differentiated
and others will remain constant F'(z)=f1'(z)f2(z)f3(z)...fn(z) + f1(z)f2'(z)f3(z)...fn(z) +
f1(z)f2(z)f3'(z)...fn(z) .................. + f1(z)f2(z)f3(z)...fn'(z) F'(z)/F(z) = [ f1'(z)f2(z)f3(z)...fn(z)
+ f1(z)f2'(z)f3(z)...fn(z) + f1(z)f2(z)f3'(z)...fn(z) .................. +
f1(z)f2(z)f3(z)...fn'(z)]/[f1(z)f2(z)f3(z)...fn(z)] F'(z)/F(z) = f1'(z)/f1(z) + f2'(z)/f2(z) +
f3'(z)/f3(z) + ...................... + fn'(z)/fn(z)](https://image.slidesharecdn.com/supposeafunctionfzisaproductofndifferentiablefunctio-230327165744-f5466bc6/85/Suppose-a-function-F-z-is-a-product-of-n-differentiable-functio-pdf-1-320.jpg)
