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प्रश्न
Find the second order derivatives of the following : e4x. cos 5x
बेरीज
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उत्तर
Let y = e4x. cos 5x
Then `"dy"/"dx" = "d"/"dx"(e^(4x).cos5x)`
= `e^(4x)."d"/"dx"(cos5x) + cos5x."d"/"dx"(e^(4x))`
= `e^(4x).(-sin5x)."d"/"dx"(5x) + cos5x xx e^(4x)."d"/"dx"(4x)`
= – e4x . sin 5x × 5 + e4x cos 5x × 4
= e4x (4 cos 5x – 5 sin 5x)
and `(d^2y)/(dx^2) = "d"/"dx"[e^(4x)(4cos5x - 5sin5x)]`
`= e^(4x)"d"/"dx"(4cos5x - 5sin5x) + (4cos5x - 5sin5x)."d"/"dx"(e^(4x))`
`= e^(4x)[4 (- sin 5x)."d"/"dx"(5x) - 5cos5x."d"/"dx"(5x)] + (4cos5x - 5sin5x) xx e^(4x)."d"/"dx"(4x)`
= e4x [– 4 sin 5x × 5 – 5 cos 5x × 5] + (4 cos 5x – 5 sin 5x) e4x × 4
= e4x(– 20 sin 5x – 25 cos 5x + 16 cos 5x – 20 sin 5x)
= e4x (– 9 cos 5x – 40 sin 5x)
= – e4x (9 cos 5x + 40 sin 5x)
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