Question

What is the derivative of y=sec^-1(2s+1)

Answer 1

\[y=\sec ^{-1}(2s+1)\]

Answer 2

\[\frac{ d }{ dx }\sec ^{-1}x=\frac{ 1 }{ x ~ \sqrt{x^2-1} }\]

Answer 3

I know that, and I got that far, but i end up with\[ \frac{ 1 }{ \left| 2s+1 \right| \sqrt{(2s)^{2}}}\]

Answer 4

The answer is supposed to be a \[\sqrt{s ^{2}+s}\]

Answer 5

\[\frac{ dy }{ ds }=\frac{ 1 }{ \left| 2s+1 \right|\sqrt{\left( 2s+1 \right)^2-1} }*2\]
\[\frac{ dy }{ ds }=\frac{ 1 }{ \left| 2s+1 \right|\sqrt{4s^2+4s+1-1} }*2\]
?

Answer 6

oh, I didn't square the (2s+1) correctly! Thank you!!

Answer 7

yw