Husseinshimaljasimdini

Name:hussein.shimal.jasim.dini Location:baghdad-iraq. Specialty:physicist.university of baghdad. http://cid-966d6907c15fe410.spaces.live.com/

Logical systems,new principles and attempt to solve collatz conjecture

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Speedy deletion of Image:MY MATH. IDEAS.pdf

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chain rule

does always dy\dx=1\(dx\dy)?thank you.Husseinshimaljasimdini (talk) 14:04, 16 April 2008 (UTC)

The proposed identity follows quickly from the chain rule. Suppose y = ƒ(x) and x = g(y). Then

${dy \over dx}=f'(x),$

and

${dx \over dy}=g'(y).$

Now, since

$g(f(x))=x,$

we have

${d \over dx}g(f(x))=1.$

But

${d \over dx}g(f(x))=g'(f(x))\cdot f'(x)$

(that's the chain rule). So

${\begin{aligned}1&{}=g'(f(x))\cdot f'(x).\\\\1&{}=g'(y)f'(x).\\\\{1 \over f'(x)}&{}=g'(y).\\\\{1 \over dy/dx}&{}={dx \over dy}.\end{aligned}}$

(Of course it doesn't work if it involves dividing by 0. At points where dy/dx is 0, dx/dy does not exist since there's a vertical tangent line.) Michael Hardy (talk) 20:22, 21 April 2008 (UTC)