How do you find the derivative of $sin x (tan x)$ $sinx(tanx)$ ?

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Answer 1 · 2018-03-29T08:34:48

$\frac{d}{d x} (sin x tan x) = cos x tan x + sin x {sec}^{2} x$ $d/dx(sinxtanx)=cosxtanx+sinxsec^2x$

After simplification $\to sin x + tan x sec x$ $->sinx+tanxsecx$

$(uv)'=u'v+uv'$

$u=sinx, v=tanx$

Therefore

$d/dx(sinxtanx)=(d/dxsinx)tanx+sinx(d/dxtanx)$

$=cosxtanx+sinxsec^2x$

We could simplify this answer a bit by using some basic trig identities:

$=cancelcosx(sinx/cancelcosx)+sinx(1/cos^2x)$

$=sinx+sinx/cosx(1/cosx)$

$=sinx+tanxsecx$

How do you find the derivative of sinx(tanx)sinx(tanx)sinx(tanx)?