Question

Find the second order derivatives of the following : e^2x . tan x

Answer 1

Let y = e^2x . tan x
Then `"dy"/"dx" = "d"/"dx"(e^(2x).tanx)`

= `e^(2x)."d"/"dx"(tanx) + tanx."d"/"dx"(e^(2x))`

= `e^(2x) xx sec^2x + tanx xx e^(2x)."d"/"dx"(2x)`

= e^2x . sec²x + e^2x . tanx × 2
= e^2x (sec²x + 2tan x)
and
`(d^2y)/(dx^2) = "d"/"dx"[e^(2x)(sec^2x + 2tanx)]`

= `e^(2x)."d"/"dx"(sec^2x + 2tanx) + (sec^2x + 2tanx)"d"/"dx"(e^(2x))`

= `e^(2x)["d"/"dx"(secx)^2 + 2"d"/"dx"(tanx)] + (sec^2x + 2tanx) xx e^(2x)."d"/"dx"(2x)`

= `e^(2x)[2secx."d"/"dx"(secx) + 2sec^2x] + (sec^2x + 2tanx)e^(2x) xx 2`

= e^2x(2 sec x . sec x tanx + 2sec²x) + 2e^2x(sec²x + 2tanx)
= 2e^2x(sec²x tanx + sec²x + sec²x + 2tanx)
= 2e^2x[sec²x(tanx + 1) + 1 + tan²x + 2tanx]
= 2e^2x[sec²x(1 + tanx) + (1 + tanx)²]
= 2e^2x[(1 + tanx) (sec²x + 1 + tanx)]
= 2e^2x[(1 + tanx) (1 + tan²x + 1 + tanx)]
= 2e^2x(1 + tanx) (2 + tanx + tan²x).