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प्रश्न
Find the nth derivative of the following : cos (3 – 2x)
उत्तर
Let y = cos (3 – 2x)
Differentiating both sides w.r.t. x, we get,
`dy/dx = d/dx [cos (3 – 2x)]`
`dy/dx = – sin(3 – 2x). d/dx (3 – 2x)`
`dy/dx = – sin(3 – 2x) × (0 – 2 × 1)`
`dy/dx = (– 2) (– sin(3 – 2x))`
`dy/dx = (– 2) cos [π/2 + (3 – 2x)]`
Again differentiating both sides w.r.t. x, we get,
`(d^2y)/(dx^2) = d/dx [(– 2) (– sin(3 – 2x)]`
`(d^2y)/(dx^2) = (– 2) d/dx [(– sin(3 – 2x)]`
`(d^2y)/(dx^2) = (– 2) [– cos (3 – 2x)] d/dx (3 – 2x)`
`(d^2y)/(dx^2) = (– 2) [– cos (3 – 2x)] × (0 – 2 × 1)`
`(d^2y)/(dx^2) = (– 2) [– cos (3 – 2x)] × (– 2)`
`(d^2y)/(dx^2) = (2)^2 [cos ((2π)/2) + (3 – 2x)]`
Again differentiating both sides w.r.t. x, we get,
`(d^3y)/(dx^3) = d/dx [(2)^2 (– cos (3 – 2x))]`
`(d^3y)/(dx^3) = (2)^2 d/dx [– cos (3 – 2x)]`
`(d^3y)/(dx^3) = (2)^2 [sin (3 – 2x)] d/dx (3 – 2x)`
`(d^3y)/(dx^3) = (2)^2 [sin (3 – 2x)] × (0 – 2 × 1)`
`(d^3y)/(dx^3) = (2)^2 [sin (3 – 2x)] (– 2)`
`(d^3y)/(dx^3) = (– 2)^3 [cos ((3π)/2) + (3 – 2x)]`
In general, the nth order derivative is given by
`(d^ny)/(dx^n) = (-2)^n cos[(nπ)/2 + (3 - 2x)]`.
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