Question

Y=tg(ax+k); y1=?........... ........

Answer 1

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