g(x)=sec(3x+1)g(x)=sec(3x+1)
First find the derivative. Use the derivative of the secantsecant function is the product of sec * tansec⋅tan functions. And use the chain rule:
For g(x)=sec(f(x))g(x)=sec(f(x)), the derivative is:
g'(x)=sec(f(x)tan(f(x))*f'(x)
(Alt notation: g(x)=secu the g'(x)=secu tanu * (du)/(dx))
g'(3x+1)=sec(3x+1)tan(3x+1)*3=3sec(3x+1) tan(3x+1)
For the second derivative we'll need the derivatives of sec, and tan, and we'll need the chain rule and the product rule.
g'(3x+1)=3 color(red)(sec(3x+1)) color(green) (tan(3x+1))
g''(x)=3[color(red)(3sec(3x+1)tan(3x+1))color(green) (tan(3x+1)) + color(red)(sec(3x+1)) color(green) (sec^2(3x+1)*3) ]
g''(x)=3[3sec(3x+1) tan^2 (3x+1) + 3sec^3(3x+1)]
To make the answer look a bit neater, either distribute the outermost 3, or factor the 3sec(3x+1)
g''(x)=9sec(3x+1)[tan^2 (3x+1) + sec^2(3x+1)]
If you prefer one trigonometric function, replace tan^2(3x+1) with sec^2 (3x+1) -1.
