- #1
Дьявол
- 365
- 0
Hello! I got one question for you.
How come that [tex](f \circ g)'(x) = f'(g(x)) g'(x)[/tex] ?
Since [tex](f \circ g)'(x)=f(g(x))'[/tex] , [tex]f'(g(x))=f'(g(x)) g'(x)[/tex]. And now we can rewrite the equation like [tex]1=g'(x)[/tex]
I don't understand that part.
Also I don't understand why the flawed proof of the chain rule is incorrect?
[tex]y'=\lim_{dx \rightarrow 0}\frac {dy}{dx} = \lim_{dx \rightarrow 0}\frac {dy} {du} \cdot\frac {du}{dx}=\lim_{du \rightarrow 0}\frac {dy}{du} \cdot \lim_{dx \rightarrow 0}\frac {du}{dx}=f'(u)\cdot u'(x)[/tex]
Thanks in advance.
Regards.
How come that [tex](f \circ g)'(x) = f'(g(x)) g'(x)[/tex] ?
Since [tex](f \circ g)'(x)=f(g(x))'[/tex] , [tex]f'(g(x))=f'(g(x)) g'(x)[/tex]. And now we can rewrite the equation like [tex]1=g'(x)[/tex]
I don't understand that part.
Also I don't understand why the flawed proof of the chain rule is incorrect?
[tex]y'=\lim_{dx \rightarrow 0}\frac {dy}{dx} = \lim_{dx \rightarrow 0}\frac {dy} {du} \cdot\frac {du}{dx}=\lim_{du \rightarrow 0}\frac {dy}{du} \cdot \lim_{dx \rightarrow 0}\frac {du}{dx}=f'(u)\cdot u'(x)[/tex]
Thanks in advance.
Regards.