method of differentâ€¦ - QuestionCove

method of differentiation question...

1 year ago

let g is the inverse function of f and f'(x) = $\frac{ x ^{10} }{ 1+x ^{2} }$ If g(2) = a , then g'(x) is = ?

1 year ago
sam (.sam.):

Okay do you have any ideas on how to solve this?

1 year ago

no..

1 year ago
sam (.sam.):

You can reverse process by integrating f'(x) to get f(x), then find $$f^{-1}(x)$$ from f(x)

1 year ago

is there any other method instead of integration becoz i only know some basics of integration...

1 year ago
MARC (marc_d):

i'm thinking it is like this..$g(x)=f'(x)$

1 year ago
MARC (marc_d):

$g(x)=\frac{ x^{10} }{ 1+x^2 }$

1 year ago
MARC (marc_d):

$Let~u=x^{10}~and~v=1+x^2$

1 year ago
MARC (marc_d):

$\frac{ du }{ dx }=10x^9$$\frac{ dv }{ dx }=2x$

1 year ago
MARC (marc_d):

in this case,u will hv to use this formula for differentiation$=\frac{ v\frac{ du }{ dx }-u \frac{ dv }{ dx } }{ v^2 }$

1 year ago
MARC (marc_d):

$g'(x)=\frac{ (1+x^2)(10x^9)-x^{10}(2x) }{ (1+x^2)^2 }$

1 year ago
OpenStudy (mww):

I'm not so sure you can find g'(x) here explicitly. You can find g'(2) for sure using inverse function theorem but not sure about g'(x)

1 year ago
OpenStudy (mww):

$g(x) = f^{-1}(x)$ $f(g(x)) = f(f^{-1}(x)) = x$ $(f(g(x))' = f'(g(x)) \times g'(x) = 1$ $g'(x) = \frac{ 1 }{ f'(g(x)) }$ From this it is possible to work out g'(2) given you know g(2). However doesn't give you much for all x though unless you know the explicit form of g(x)

1 year ago
ganeshie8 (ganeshie8):

so you have $g'(x) = \dfrac{1+g^2(x)}{g^{10}(x)}$

1 year ago
ganeshie8 (ganeshie8):

That doesn't look bad at all as you can find $$g'$$ whenever you knew $$g$$.

1 year ago