How do you find the second derivative of y = x^4 tan x?

Feb 21, 2015

$y ' ' = 2 {x}^{2} \left(4 x + 6 \tan x + 4 x {\tan}^{2} x + {x}^{2} {\tan}^{2} x + {x}^{2} {\tan}^{3} x\right)$.

With the pruduct rule that says:

$y = f \left(x\right) \cdot g \left(x\right) \Rightarrow y ' = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) g ' \left(x\right)$

Than:

$y ' = 4 {x}^{3} \tan x + {x}^{4} \left(1 + {\tan}^{2} x\right)$

or

$y ' = 4 {x}^{3} \tan x + {x}^{4} \cdot \frac{1}{\cos} ^ 2 x$.

From the first:

$y ' ' = 12 {x}^{2} \tan x + 4 {x}^{3} \left(1 + {\tan}^{2} x\right) + 4 {x}^{3} \left(1 + {\tan}^{2} x\right) + {x}^{4} \cdot 2 \tan x \cdot \left(1 + {\tan}^{2} x\right) =$

$= 2 {x}^{2} \left(6 \tan x + 2 x \left(1 + {\tan}^{2} x\right) + 2 x \left(1 + {\tan}^{2} x\right) + {x}^{2} \tan x \left(1 + {\tan}^{2} x\right)\right) =$

$= 2 {x}^{2} \left(6 \tan x + 2 x + 2 x {\tan}^{2} x + 2 x + 2 x {\tan}^{2} x + {x}^{2} \tan x + {x}^{2} {\tan}^{3} x\right) =$

$= 2 {x}^{2} \left(4 x + 6 \tan x + 4 x {\tan}^{2} x + {x}^{2} {\tan}^{2} x + {x}^{2} {\tan}^{3} x\right)$.