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प्रश्न
Differentiate the following w.r.t. x : `cot^-1((4 - x - 2x^2)/(3x + 2))`
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उत्तर
Let y = `cot^-1((4 - x - 2x^2)/(3x + 2))`
= `tan^-1((3x + 2)/(4 - x - 2x^2)) ...[∵ cot^-1 x = tan^-1(1/x)]`
= `tan^-1[(3x + 2)/(1 - (2x^2 + x - 3))]`
= `tan^-1 [((2x + 3) + (x - 1))/(1 - (2x + 3)(x - 1))]`
= tan–1(2x + 3) + tan–1(x – 1)
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[tan^-1(2x + 3) + tan^-1(x - 1)]`
= `"d"/"dx"[tan^-1(2x + 3)] + "d"/"dx"[tan^-1(x - 1)]`
= `(1)/(1 + (2x + 3)^2)."d"/"dx"(2x + 3) + (1)/(1 + (x - 1)^2)."d"/"dx"(x - 1)`
= `(1)/(1 + (2x + 3)^2).(2 xx 1 + 0) + (1)/(1 + (x - 1)^2).(1 - 0)`
= `(2)/(1 + (2x + 3)^2) + (1)/(1 + (x - 1)^2`.
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