Ok... Our class (as our homework assignment) needs to find the following derivatives using implicit..

 

[math]sin^{-1}(x), cos^{-1}(x), tan^{-1}(x), cot^{-1}(x), csc^{-1}(x), sec^{-1}(x)[/math]

 

and, since our book doesn't give these derivatives, I don't know if I got them right (because we have to then use them in the assignment...)

 

[math]sin^{-1}(x)=y[/math]

 

[math]sin(sin^{-1}(x))=sin(y)[/math]

 

[math]x=sin(y)[/math]

 

[math]1=cos(y)y'[/math]

 

[math]y'=\frac{1}{cos(y)}=sec(y)[/math]

 

using triangles and trig identities, since arcsin(x)=y, y is the angle with x the side opposite... so x/1 would be opposite over hypotinuse... so the last side is

 

[math]\sqrt{1^2-x^2}[/math]

 

so, sec is the reciprical of the cos, so it's hypotinuse over adjecent...

 

[math]sec(y)=\frac{1}{\sqrt{1-x^2}}[/math]

 

and back subsititution...

 

[math]y'=sec(y)=\frac{1}{\sqrt{1-x^2}}[/math]

 

Then, to generize it using the chain rule...

 

[math]y'=sec(y)=\frac{1}{\sqrt{1-x^2}}=\frac{u'}{\sqrt{1-u^2}}[/math]

 

Is this right? And does the same process work for the others?

 

Let's see... using the process, this is what I get for the inverse trig derivatives...

 

[math]y=sin^{-1}(x), y'=\frac{u'}{\sqrt{1-u^2}}[/math]

 

[math]y=cos^{-1}(x), y'=-\frac{u'}{\sqrt{1-u^2}}[/math]

 

[math]y=tan^{-1}(x), y'=\frac{u'}{1+u^2}[/math]

 

[math]y=cot^{-1}(x), y'=-\frac{u'}{1+u^2}[/math]

 

[math]y=sec^{-1}(x), y'=\frac{u'}{u\sqrt{u^2-1}}[/math]

 

[math]y=csc^{-1}(x), y'=-\frac{u'}{u\sqrt{u^2-1}}[/math]

 

anyone notice an error here? or did I do it right?

If your final answers are correct, shouldn't that be in itself a good check?

It certainly looks to be the correct approach.

If your final answers are correct, shouldn't that be in itself a good check?

 

except that I have no way of knowing what the final answers are... the book doesn't have them.

Google is your friend.

 

Just note that arcsin = sin^-1, arccos = cos^-1 etc.