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Question
Find the second order derivatives of the following : e2x . tan x
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Solution
Let y = e2x . 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)`
= e2x . sec2x + e2x . tanx × 2
= e2x (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`
= e2x(2 sec x . sec x tanx + 2sec2x) + 2e2x(sec2x + 2tanx)
= 2e2x(sec2x tanx + sec2x + sec2x + 2tanx)
= 2e2x[sec2x(tanx + 1) + 1 + tan2x + 2tanx]
= 2e2x[sec2x(1 + tanx) + (1 + tanx)2]
= 2e2x[(1 + tanx) (sec2x + 1 + tanx)]
= 2e2x[(1 + tanx) (1 + tan2x + 1 + tanx)]
= 2e2x(1 + tanx) (2 + tanx + tan2x).
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