OpenStudy (purpleshadow):

method of differentiation question...

1 year ago
OpenStudy (purpleshadow):

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
OpenStudy (purpleshadow):

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
OpenStudy (purpleshadow):

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