Let y=e2xtanx.
Step 1 — first derivative by the product rule: dxdy=2e2xtanx+e2xsec2x=e2x(2tanx+sec2x).
Step 2 — differentiate again, treating e2x and (2tanx+sec2x) as the two factors. First note dxd(2tanx+sec2x)=2sec2x+2secx⋅secxtanx=2sec2x+2sec2xtanx.
dx2d2y=2e2x(2tanx+sec2x)+e2x(2sec2x+2sec2xtanx)
Step 3 — expand and collect: =e2x[4tanx+2sec2x+2sec2x+2sec2xtanx]=e2x[4tanx+4sec2x+2sec2xtanx].
dx2d2y=2e2x(2tanx+2sec2x+sec2xtanx)
✓Final answer
dx2d2y=2e2x(2tanx+2sec2x+sec2xtanx)