Find the second derivative of f(x)=x^4 tan x?

$f ' ' \left(x\right) = 2 {x}^{4} \setminus {\sec}^{2} x \setminus \tan x + 8 {x}^{3} \setminus {\sec}^{2} x + 12 {x}^{2} \setminus \tan x$

Explanation:

Given that

$f \left(x\right) = {x}^{4} \setminus \tan x$

Differentiating above function w.r.t. $x$ using product rule as follows

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{4} \setminus \tan x\right)$

$f ' \left(x\right) = {x}^{4} \frac{d}{\mathrm{dx}} \setminus \tan x + \setminus \tan x \frac{d}{\mathrm{dx}} {x}^{4}$

$= {x}^{4} \setminus {\sec}^{2} x + \setminus \tan x \left(4 {x}^{3}\right)$

$= {x}^{4} \setminus {\sec}^{2} x + 4 {x}^{3} \setminus \tan x$

Differentiating above equation w.r.t. $x$ using product rule as follows

$\frac{d}{\mathrm{dx}} f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{4} \setminus {\sec}^{2} x + 4 {x}^{3} \setminus \tan x\right)$

$f ' ' \left(x\right) = {x}^{4} \frac{d}{\mathrm{dx}} \left({\sec}^{2} x\right) + \setminus {\sec}^{2} x \frac{d}{\mathrm{dx}} \left({x}^{4}\right) + 4 {x}^{3} \frac{d}{\mathrm{dx}} \left(\setminus \tan x\right) + 4 \setminus \tan x \frac{d}{\mathrm{dx}} \left({x}^{3}\right)$

$= {x}^{4} \left(2 \setminus \sec x \setminus \sec x \setminus \tan x\right) + \setminus {\sec}^{2} x \left(4 {x}^{3}\right) + 4 {x}^{3} \left(\setminus {\sec}^{2} x\right) + 4 \setminus \tan x \left(3 {x}^{2}\right)$

$= 2 {x}^{4} \setminus {\sec}^{2} x \setminus \tan x + 4 {x}^{3} \setminus {\sec}^{2} x + 4 {x}^{3} \setminus {\sec}^{2} x + 12 {x}^{2} \setminus \tan x$

$= 2 {x}^{4} \setminus {\sec}^{2} x \setminus \tan x + 8 {x}^{3} \setminus {\sec}^{2} x + 12 {x}^{2} \setminus \tan x$