Question #ec712

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Answer 1 · 2017-04-07T12:43:14

$d/dxtan^2x=2tanx*sec^2x$

$d/dxsec^2x=2sec^2xtanx$

$d/dxtan^2x$

$r(u)^(r-1)*d/(du)$

= $2tanx*d/dx(tanx)$

$d/dxtanx=sec^2x$ (common derivative)

Therefore the final answer is,

$d/dxtan^2x=2tanx*sec^2x$

$d/dxsec^2x$

Apply chain rule

$r(u)^(r-1)*d/(du)$

$2secx*d/(dx)secx$

$d/(dx)secx=secx*tanx$ (common derivative)

Therefore the final answer is,

$d/dxsec^2x= 2secx*secx*tanx*$

Which simplifies to,

$d/dxsec^2x=2sec^2xtanx$