`(d)/(dx) {e ^(ax)sin (bx+c)}=`

Show that (d)/(dx) e^(ax) cos (bx + c) = r e^(ax) cos (bx + c + alpha) where r= sqrt(a^(2) + b^(2)), cosalpha= (a)/(r ), sin alpha = (b)/(r ) and (d^(2))/(dx^(2)) e^(ax) cos (ax + c) = r^(2) e^(ax ) cos (bx + c + 2 alpha) .

d/(dx) (e^(x sin x)) =

d/(dx) (e^(x sin x)) =

(d) / (dx) e ^ ((x sin x + cos x))

(d)/(dx)[e^(x)sin sqrt(3)x]=

(d)/(dx)(e^(x)sin sqrt(3)x) equals-

(d)/(dx) (e^(5x)) = …….

(d)/(dx){log_(e)(ax)^(x)}

d/dx (e^(sin√x))