Question #ec712

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Answer 1 · 2017-04-07T12:43:14

$\frac{d}{d x} {tan}^{2} x = 2 tan x \cdot {sec}^{2} x$ $d/dxtan^2x=2tanx*sec^2x$

$\frac{d}{d x} {sec}^{2} x = 2 {sec}^{2} x tan x$ $d/dxsec^2x=2sec^2xtanx$

$\frac{d}{d x} {tan}^{2} x$ $d/dxtan^2x$

$r {(u)}^{r - 1} \cdot \frac{d}{d u}$ $r(u)^(r-1)*d/(du)$

= $2 tan x \cdot \frac{d}{d x} (tan x)$ $2tanx*d/dx(tanx)$

$\frac{d}{d x} tan x = {sec}^{2} x$ $d/dxtanx=sec^2x$ (common derivative)

Therefore the final answer is,

$\frac{d}{d x} {tan}^{2} x = 2 tan x \cdot {sec}^{2} x$ $d/dxtan^2x=2tanx*sec^2x$

$\frac{d}{d x} {sec}^{2} x$ $d/dxsec^2x$

Apply chain rule

$r {(u)}^{r - 1} \cdot \frac{d}{d u}$ $r(u)^(r-1)*d/(du)$

$2 sec x \cdot \frac{d}{d x} sec x$ $2secx*d/(dx)secx$

$\frac{d}{d x} sec x = sec x \cdot tan x$ $d/(dx)secx=secx*tanx$ (common derivative)

Therefore the final answer is,

$\frac{d}{d x} {sec}^{2} x = 2 sec x \cdot sec x \cdot tan x \cdot$ $d/dxsec^2x= 2secx*secx*tanx*$

Which simplifies to,

$\frac{d}{d x} {sec}^{2} x = 2 {sec}^{2} x tan x$ $d/dxsec^2x=2sec^2xtanx$