Question

y=tg(ax+k); y1=?
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Answer 1

$(tgx)'= frac{1}{cos^2x} quad to quad (tgu)'= frac{1}{cos^2u}cdot u'\\\y=tg(ax+k)\\y'= frac{1}{cos^2(ax+k)}cdot (ax+k)'= frac{1}{cos^2(ax+k)} cdot a\\\P.S.; ; (ax+k)'=(ax)'+k'=acdot x'+0=acdot 1=a; ; (a,k=const)$