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प्रश्न
Find the nth derivative of the following : eax+b
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उत्तर
Let y = eax+b
Then `"dy"/"dx" = "d"/"dx"(e^(ax + b))`
= `e^(ax + b)."d"/"dx"(ax + b)`
= `e^(ax + b) xx (a xx 1 + 0)`
= aeax+b
`(d^2y)/(dx^3) = "d"/"dx"(ae^(ax + b))`
= `a."d"/"dx"(ax + b)`
= `ae^(ax + b) xx (a xx 1 + 0)`
= a2.eax+b
`(d^3y)/(dx^3) = "d"/"dx"[a^2e^(ax + b)]`
= `a^2"d"/"dx"(e^(ax + b))`
= `a^2e^(ax + b)."d"/"dx"(ax + b)`
= a2eax+b x (a x 1 + 0)
= a3.eax+b
In genaral, the nth order derivative is given by
`(d^ny)/(dx^n)` = an . eax+b.
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