1. differentia secx????????

why is the differential of secx tanx when secx=/cosx=tanx/sinx

im also confused as to why when you've got d(sec^(2)x)/dx)

you can't do sec^(2)x=cos^(-2)x
therefore dy/dx = 2cosxsinx=sin2x

i think its wrong to say 1/cos^(2)x=cos^(-2)x but im not sure, is this the same as 1/arccosx im so confused,,,,,,,,,,,,,,,,,,,,,

genreally just want to know how to diff secx >>>???? thnx,,,,but expaination as to y im wrong would be much appreciated thnx

2. Originally Posted by i_zz_y_ill
why is the differential of secx tanx when secx=/cosx=tanx/sinx

im also confused as to why when you've got d(sec^(2)x)/dx)

you can't do sec^(2)x=cos^(-2)x
therefore dy/dx = 2cosxsinx=sin2x

i think its wrong to say 1/cos^(2)x=cos^(-2)x but im not sure, is this the same as 1/arccosx im so confused,,,,,,,,,,,,,,,,,,,,,

genreally just want to know how to diff secx >>>???? thnx,,,,but expaination as to y im wrong would be much appreciated thnx
Your post is difficult to read and follow. It would benefit from better setting out.

To differentiate y = sec x = 1/cos x, use the quotient rule.

3. Hi
Originally Posted by i_zz_y_ill
why is the differential of secx tanx when secx=/cosx=tanx/sinx
I'm not sure to have understood what you mean. Anyway :

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(\sec x)=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{\c os x}\right) =-\frac{-\sin x}{\cos^2x}=\sec x \tan x$

im also confused as to why when you've got d(sec^(2)x)/dx)

you can't do sec^(2)x=cos^(-2)x
therefore dy/dx = 2cosxsinx=sin2x
Because $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left( u^n (x)\right)=n\cdot u'(x)\cdot u^{n-1}(x)$ which is not what you wrote. (-2-1=-3 )