Question

Find derivative of tanx w.r.t.x​

Answer 1

Answer:

Tanx = - cotx

Step-by-step explanation:

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Answer 2

Answer:

${sec}^{2} x$

Step-by-step explanation:

$(\frac{d}{dx} ) \: tanx \: = ( \frac{d}{dx} ) \: ( \frac{cosx}{sinx} ) \\ it \: is \: in \: the \: from \frac{u}{v} and \: the \: derivate \: is \: \\ \frac{vdu - udu}{ {v}^{2} } where \: u \: = sinx \: v = cosx \\ (\frac{d}{dx} ) \: ( \frac{sinx}{cosx} ) \: = cosx.cosx \\ - \times \frac{sinx.( - sinx)}{ {cos}^{2}x } \\ \frac{ {cos}^{2} x + {sin}^{2}x }{ {cos}^{2} x} = \frac{1}{ {cos}^{2} x} = {sec}^{2} x \: as \\ {cos}^{2} x + {sin}^{2} x = 1$

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