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Differentiate the following w.r.t. x : tan-1(5-x6x2-5x-3) - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t. x :

`tan^-1((5 -x)/(6x^2 - 5x - 3))`

योग
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उत्तर

Let `y = tan^-1((5 -x)/(6x^2 - 5x - 3))`

`y = tan^-1((5 -x)/(6x^2 - 5x - 4 + 1))`

`y = tan^-1[(5 - x)/(1 + (6x^2 - 5x - 4))]`

`y = tan^-1[((2x + 1) - (3x - 4))/(1 + (2x + 1)(3x - 4))]`

`y = tan^-1(2x + 1) – tan^-1(3x  –  4)      ...[tan^(-1) x - tan^(-1) y = tan^(-1) ((x - y)/(1 + xy))]`

Differentiating w.r.t. x, we get,

`dy/dx = d/dx [tan^-1(2x + 1)  – tan^-1(3x  –  4)]`

`dy/dx = d/dx [tan^-1(2x + 1)] - d/dx [tan^-1(3x - 4)]`

`dy/dx = (1)/(1 + (2x + 1)^2). d/dx (2x + 1) - (1)/(1 + (3x - 4)^2). d/dx (3x - 4)  ...[tan^(-1) x = 1/(1 + x^2)]`

`dy/dx = (1)/(1 + (2x + 1)^2).(2 xx 1 + 0) - (1)/(1 + (3x - 4)^2).(3 xx 1 - 0) ...[(d/dx x = 1), (d/dx k = 0)]`

`dy/dx = (2)/(1 + (2x + 1)^2) - (3)/(1 + (3x - 4)^2`.

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Differentiation
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Differentiation - Exercise 1.2 [पृष्ठ ३०]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.2 | Q 10.8 | पृष्ठ ३०

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