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प्रश्न
Find the nth derivative of the following:
y = e8x . cos (6x + 7)
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उत्तर
y = e8x . cos (6x + 7)
∴ `"dy"/"dx" = "d"/"dx"[e^(8x).cos (6x + 7)]`
= `e^(8x)."d"/"dx"[cos (6x + 7)] + cos (6x + 7)."d"/"dx"(e^(8x))`
= `e^(8x).[-sin(6x + 7)]."d"/"dx"(6x + 7) + cos(6x + 7).e^(8x)."d"/"dx"(8x)`
= – e8x sin (6x + 7) x (b x 1 + 0) + e8xcos(6x + 7) x a x 1
= e8x [a cos (6x + 7) – b sin (6x + 7)]
= `e^(8x).sqrt(a^2 + b^2)[a/sqrt(a^2 + b^2)cos(6x + 7) - b/sqrt(a^2 + b^2)sin(6x + 7)]`
Let `a/sqrt(a^2 + b^2) = cos x and b/sqrt(a^2 + b^2) = sin x`
Then tan ∞ = `b/a`
∴ ∞ = `tan^-1(b/a)`
∴ `"dy"/"dx" = e^(8x).sqrt(a^2 + b^2)[cosoo.cos(bx + c) - sinoo.sin(bx + c)]`
= `e^(8x).(a^2 + b^2)^(1/2).cos(6x + 7 + x)`
`(d^2y)/(dx^2) = "d"/"dx"[e^(8x).(a^2 + b^2)^(1/2).cos(6x + 7 + oo)]`
= `(a^2 + b^2)^(1/2)."d"/"dx"[e^(8x).cos(6x + 7 + oo)]`
= `(a^2 + b^2)^(1/2)[e^(8x)."d"/"dx"{cos(6x + 7 + oo)} + cos(6x + 7 + oo)."d"/"dx"(e^(8x))]`
= `(a^2 + b^2)^(1/2)[e^(8x).{-sin(6x + 7 + oo)}."d"/"dx"(6x + 7 + oo) + cos(6x + 7 + oo).e^(8x)."d"/"dx"(8x)]`
= `(a^2 + b^2)^(1/2)[-e^(8x)sin(6x + 7 + oo) xx (b xx 1 + 0 + 0) + cos(6x + 7 + oo).e^(8x) xx a xx 1]`
= `e^(8x).(a^2 + b^2)^(1/2)[a cos (6x + 7 + oo) - bsin(6x + 7 + oo)]`
= `e^(8x).(a^2 + b^2)^(1/2).sqrt(a^2 + b^2)[a/sqrt(a^2 + b^2)cos(6x + 7 + oo) = b/sqrt(a^2 + b^2)sin(6x + 7 + oo)]`
= `e^(8x).(a^2 + b^2)^(2/2)[cosoo.cos(6x + 7 + ∞) - sinoo.sin(6x + 7 + oo)`
= `e^(8x).(a^2 + b^2)^(2/2).cos(6x + 7 + oo + oo)`
= `e^(8x).(a^2 + b^2)^(2/2).cos(6x + 7 + 2oo)`
Similarly.
`(d^3y)/(dx^3) = e^(8x).(a^2 + b^2)^(3/2).cos(6x + 7 + 3oo)`
In general, the nth order derivative is given by
`(d^ny)/(dx^n) = e^(8x).(a^2 + b^2)^(n/2).cos(6x + 7 + noo)`,
Where ∞ = `tan^-1(b/a)`
∴ `(d^ny)/(dx^n) = e^(8x).(10)^n.cos[6x + 7 + ntan^-1(3/4)]`
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