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Find f'(x) for f(x) = arcsec(x/2)
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Substitute u = x/2 into this equation: \[D _{x} \sec^-1 u = \frac{ 1 }{ u \sqrt{u^2-1} }D _{x} u\]
I hope that helps
That would make it \[\frac{ 1 }{ \frac{ x }{ 2 } \sqrt{(\frac{ x }{ 2 })^{2}-1}} \times \frac{ 1 }{ 2 }\] right? I'm just not sure how to go about simplifying this.
\[\frac{ 1 }{ \frac{ x }{2 } \times \frac{ 1 }{ 2 } \sqrt{x^2 - 4}} \times \frac{ 1 }{ 2 }\]
Then the 1/2 on the right cancels out with the 1/2 next to the radical
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Ah, alright. Thank you very much.
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