inverse (1 Viewer)

perseverance10 · Apr 29, 2010

just having a few problems with finding the derivative of the following:

1. f(x)=inverse tan px

and

2. f(x)=inverse cos (1-x^2)/(1+x^2)

any help would be greatly appreciated =]

thanks in advance

Drongoski · Apr 29, 2010

A quick attempt yields the following. But am not sure if they are correct. Do you have the answers ?

$\frac{\mathrm{d} }{\mathrm{d} x}tan^{-1}\pi x\ =\ \frac{\pi }{1+(\pi x)^{2}}\\ \\ \\ \frac{\mathrm{d} }{\mathrm{d} x}cos^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right )\ \ =\ \ \frac{2}{1+x^{2}}$

If above correct I can post solution.

perseverance10 · Apr 30, 2010

Drongoski said:
A quick attempt yields the following. But am not sure if they are correct. Do you have the answers ?

$\frac{\mathrm{d} }{\mathrm{d} x}tan^{-1}\pi x\ =\ \frac{\pi }{1+(\pi x)^{2}}\\ \\ \\ \frac{\mathrm{d} }{\mathrm{d} x}cos^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right )\ \ =\ \ \frac{2}{1+x^{2}}$

If above correct I can post solution.

hey yeh they're correct except instead of pie it's just the letter 'p' but u got it right regardless

what was your 'a' in 1?

Drongoski · Apr 30, 2010

what was your 'a' in 1?

don't understand.

perseverance10 · Apr 30, 2010

'a' is in the formula for differentiating inverse functions --> on the standard integral sheet =]

and 1 as in question 1

could u pls show me your working? i don't understand =/

Drongoski · Apr 30, 2010

inverse (1 Viewer)

perseverance10

New Member

Drongoski

Well-Known Member

perseverance10

New Member

Drongoski

Well-Known Member

perseverance10

New Member

Drongoski

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)