Evaluating $g'''(\pi/4)$, given $g(x) = \sec (x)$ Given the function $$g(x)=\sec(x)$$, I have to solve for $g'''(\pi /4)$.
I calculated the 3rd derivative to be $$ g'''(x)=\sec x\tan ^3 x+5\sec ^3x \tan x$$
I just don't know how to evaluate the function for a value in terms of pi.
 A: To finish up this problem, you need only plug in a value for $x$. You know what $\tan(\pi/4)$ and $\sec(\pi/4)$ are, right? Then evaluate the expression that Dr. Graubner (and you) wrote down!
One difficulty might be in the phrasing: we don't usually say "solve for" unless some variable is unknown, as in "Solve for $x$ in 
$$
x^2 = x + 1."
$$
When we know the values of everything in an expression, we tend to say "Simplify" or "evaluate", as in 
"Evaluate 
$$
\cos^2(x) - (1-\sin^2(x)) + 2 \sin(x)
$$
when $x = 0$."
The solution to this latter problem is just
$$
\cos^2(0) - (1-\sin^2(0)) + 2 \sin(0) = 1^2 - (1-0^2) + 2\cdot 0 = 0. 
$$
A: You just plug $x=\pi/4$ into the function you found for $g'''(x)$ so: $$g'''(\pi/4)=5\tan(\pi/4) \sec^3 (\pi/4) + \sec (\pi/4) \tan^3(\pi/4) = 5(1)(\sqrt 2)^3 + (\sqrt2)(1)^3 = 5(2\sqrt2) + \sqrt2 = 11\sqrt2$$
A: You've done just fine. To evaluate at $\pi/4$, it may help if we simplify $g'''(x)$ first.
We can factor out the common factors $\sec(x)$ and $\tan(x)$, and use the identity $$\tan^2x + 1 = \sec^2 x \iff \tan^2 x = \sec^2 x - 1$$ Then we get $$g'''(x) = \sec(x)\tan(x)\Big(5\sec^2 x + \underbrace{\tan^2(x)}_{=\sec^2 (x) - 1}\Big) = \sec(x)\tan(x)\Big(6\sec^2(x) - 1\Big)$$
Now, plug in $x = \frac{\pi}{4}$ and evaluate.
$$\sec\left(\frac \pi 4\right) = \sqrt 2\;\text{ and }\;\;\tan\left(\frac \pi 4\right) = 1$$
This gives us $$g'''(x) = \sqrt 2(1)\Big(6(\sqrt 2)^2 - 1\Big) = \sqrt 2(12- 1) = 11\sqrt 2$$
A: it can factorized to this here
$-\frac{1}{2} (\cos (2 x)-11) \tan (x) \sec ^3(x)$
