Question

How would you find the derivative of y = sin−1(6x + 1)?

Answer 1

if \[y=\sin^{-1}(6x+1) => \sin(y)=6x+1 => y' \cos(y)=6 => y'=\frac{6}{\cos(y)}\]
if sin(y)=(6x+1)/(1) (=opp/hyp)

Answer 2

\[\cos(y)=\frac{\sqrt{1^2-(6x+1)^2}}{1} (=\frac{adj}{hyp})\]

Answer 3

\[y'=\frac{6}{\sqrt{1-(6x+1)^2}}\]

Answer 4

thanks!

Answer 5

i tried applying that to another question and this is how far I got:
\[y=25\arctan \sqrt{x}\]
\[y=25\tan^{-1} (\sqrt{x})\]
\[25\tan(y)=\left(\begin{matrix}\sqrt{x} \\ 25\end{matrix}\right)\]
\[(\sec(y))(\tan(y))=\left(\begin{matrix}\sqrt{x} \\ 25\end{matrix}\right)\]

Answer 6

what would i do next?

Answer 7

\[y=25\tan^{-1}(\sqrt{x}) => \frac{y}{25}=\tan^{-1}(\sqrt{x})\]
\[=>\tan(\frac{y}{25})=\sqrt{x}=\frac{\sqrt{x}}{1} (=\frac{opp}{adj})\]