how do you differentiate inverse cosecx (1 Viewer)

[WEMG]

how do you differentiate inverse cosecx

 

[hscishard]

Let y= inverse cosecx
x=1/siny
Find dx/dy, flip it. Sub y=inverse cosecx
I think thats how you do it..

EDIT:OH, and you'll need to play around with the rt angled triangles(hyp is 1, opp is x,etc)

 

[Trebla]

Note that technically



but the range is defined for this function as



which means that cos y > 0 hence the positive root is taken

 

[Drongoski]

hscishard has the correct general idea. But I doubt you'd ever be asked to do it in MX1.

Let y = cosec^-1x ==> x = 1/(sin y) ==> sin y = 1/x and cos y = sqrt(1 - sin^2 y)



Now: differentiating wrt x the equation: 1/sin y = x ==> - cos y/sin^2 y . (dy/dx) = 1

==> dy/dx = - sin²y / cos y =- (1/x)² . x/(sqrt(x² - 1))

= - 1/[x sqrt(x² -1)]

 

[Drongoski]

Ah! Trebla's LaTeX version looks much nicer.

 

[hscishard]

Ahh (1-sin^2)^0.5 eliminates the need to use rt angle triangles.
i like that

 

[WEMG]

Yep I got it, thnks guys

 

[cutemouse]

This would usually come under 'harder 3U' in MX2.

 

[iSplicer]

This would usually come under 'harder 3U' in MX2.

Oh how I wished that were true at 10am on the 27th of October 2010.

 

[mitchy_boy]

This would usually come under 'harder 3U' in MX2.

Really? I can't see why, it's nothing too tricky...

 

[Bored Of Fail]

Really? I can't see why, it's nothing too tricky...

its the stuff you do in first semester math at uni

the method itself is not that hard but most people doing the hsc wouldnt have seen this method before

 

[Trebla]

Inverse reciprocal trigonometric functions are not in the Ext1 course.