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प्रश्न
Find the nth derivative of the following : sin (ax + b)
उत्तर
Let y = sin (ax + b)
Then `"dy"/"dx" = "d"/"dx"[sin(ax + b)]`
= `cos(ax + b)."d"/"dx"(ax + b)`
= cos (ax + b) x (a x 1 + 0)
= `asin[pi/2 + (ax + b)]`
`(d^2y)/(dx^2) = "d"/"dx"[acos(ax + b)]`
= `a"d"/"dx"[cos (ax + b)]`
= `a[- sin(ax + b)]."d"/"dx"(ax + b)`
= a[ – sin (ax + b)] x (a x 1 + 0)
= a2 . sin[π + (ax + b)]
= `a^2.sin[(2pi)/2 + (ax + b)]`
`(d^3y)/(dx^3) = "d"/"dx"[-a^2sin(ax + b)]`
= `-a^2"d"/"dx"[sin(ax + b)]`
= `-a^2.cos(ax + b)."d"/"dx"(ax + b)`
= – a2. cos(ax + b) x (a x 1 + 0)
= `a^3.sin[(3pi)/2 + (ax + b)]`
In general, the nth order derivative is given by
`(d^ny)/(dx^n) = a^n.sin[(npi)/2 + (ax + b)]`.
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