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Differentiate the following w.r.t.x: 1+sinx°1-sinx° - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t.x: `(1 + sinx°)/(1 - sinx°)`

योग
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उत्तर

Let y = `(1 + sinx°)/(1 − sinx°)` 

y = `(1 + sin((πx)/180))/(1 − sin((πx)/180))          ...[∵ x° = ((pix)/180)^°]`

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

`dy/dx = d/dx [(1 + sin((πx)/180))/(1 − sin((πx)/180))]`

`dy/dx = ([1 − sin((πx)/180)]. d/dx [1 + sin((πx)/180)] − [1 + sin((πx)/180)]. d/dx [1 − sin((πx)/180)])/[1 − sin((πx)/180)]^2`

`dy/dx = ([1 − sin((πx)/(180))].[0 + cos((πx)/(180)). d/dx ((πx)/(180)) - [1 + sin((πx)/(180))].[0 − cos((πx)/(180)). d/dx ((πx)/(180))]))/[1 − sin((πx)/180)]^2`

`dy/dx = ((1 − sinx°)[(cosx°) × π/(180) × 1] - (1 + sinx°)[(− cosx°) × π/(180) × 1])/(1 − sinx°)^2`

`dy/dx = (π/(180)cosx°(1 − sinx° +  1 + sinx°))/(1 - sinx°)^2`

`dy/dx = (πcosx°)/(90(1 − sinx°)^2`.

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अध्याय 1: Differentiation - Exercise 1.1 [पृष्ठ १२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.1 | Q 3.08 | पृष्ठ १२

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