How do you find second derivative of #g(x)=sec(3x+1)#?

Question

Cameron Alexander · Accepted Answer

#g(x)=sec(3x+1)# First find the derivative.  Use the derivative of the #secant# function is the product of #sec * tan# functions.  And use the chain rule:   For #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#.

Charles Anctil · Answer

undefined To find the second derivative of $ g(x) = \sec(3x+1) $, follow these steps:

1. Find the first derivative of $ g(x) $ using the chain rule.
2. Then, find the second derivative by differentiating the result obtained in step 1.

First derivative of $ g(x) $:

$$ g'(x) = \sec(3x+1) \tan(3x+1) \cdot 3 $$

Second derivative of $ g(x) $:

$$ g''(x) = \frac{d}{dx} \left( \sec(3x+1) \tan(3x+1) \cdot 3 \right) $$

$$ g''(x) = 3 \cdot \sec(3x+1) \tan(3x+1) \cdot \sec^2(3x+1) \cdot 3 $$

$$ g''(x) = 9 \sec(3x+1) \tan(3x+1) \sec^2(3x+1) $$