Answer
\[y'' = 2\,\left( {{{\sec }^2}x\tan x + {{\csc }^2x}\cot x} \right)\]
Work Step by Step
\[\begin{gathered}
y = \sec x\csc x \hfill \\
\hfill \\
differentiate\,\,\,to\,\,find\,\,{y^{\,,}} \hfill \\
use\,\,the\,\,product\,\,rule \hfill \\
\hfill \\
y' = \csc x\,\left( {\sec x\tan x} \right) + \sec x\,\left( { - \csc x\cot x} \right) \hfill \\
\hfill \\
{\text{use}}\;\;{\text{trigonometric identities}} \hfill \\
\hfill \\
y' = \frac{1}{{\sin x}}\,\left( {\sec x\frac{{\sin x}}{{\cos x}}} \right) - \frac{1}{{\cos x}}\,\left( {\csc x\frac{{\cos x}}{{\sin x}}} \right) \hfill \\
\hfill \\
{\text{Simplify}} \hfill \\
\hfill \\
y' = {\sec ^2}x - {\csc ^2}x \hfill \\
\hfill \\
differentiate\,\,\,to\,\,find\,\,{y^{\,,,}} \hfill \\
\hfill \\
y'' = 2{\sec ^2}x\tan x + 2{\csc ^2}x\cot x \hfill \\
\hfill \\
y'' = 2\,\left( {{{\sec }^2}x\tan x + {{\csc }^2x}\cot x} \right) \hfill \\
\end{gathered} \]
