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Differentiate the following w.r.t. x : cot-1[1+sin (4x3)+1-sin (4x3)1+sin (4x3)-1-sin (4x3)] - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t. x :

`cot^-1[(sqrt(1 + sin  ((4x)/3)) + sqrt(1 - sin  ((4x)/3)))/(sqrt(1 + sin  ((4x)/3)) - sqrt(1 - sin  ((4x)/3)))]`

योग
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उत्तर

Let y = `cot^-1[(sqrt(1 + sin  ((4x)/3)) + sqrt(1 - sin  ((4x)/3)))/(sqrt(1 + sin  ((4x)/3)) - sqrt(1 - sin  ((4x)/3)))]`

= `1 + sin ((4x)/3)`

= `1 + cos(pi/2 - (4x)/3)`

= `2cos^2(pi/4 - (2x)/3)`

∴ `sqrt(1 + sin((4x)/3)) = sqrt(2)cos(pi/4 - (2x)/3)`

Also, `1 - sin ((4x)/3)`

= `1 - cos(pi/2 - (4x)/3)`

= `2sin^2(pi/4 - (2x)/3)`

∴ `sqrt(1 - sin((4x)/3)) = sqrt(2)sin(pi/4 - (2x)/3)`

∴ `(sqrt(1 + sin  ((4x)/3)) + sqrt(1 - sin ((4x)/3)))/(sqrt(1 + sin((4x)/3) - sqrt(1 - sin((4x)/3)`

= `(sqrt(2)cos(pi/4 - (2x)/3) + sqrt(2)sin(pi/4 - (2x)/3))/(sqrt(2)cos(pi/4 - (2x)/3) - sqrt(2)sin(pi/4 - (2x)/3)`

= `(cos(pi/4 - (2x)/3) + sin(pi/4 - (2x)/3))/(cos(pi/4 - (2x)/3) - sin(pi/4 - (2x)/3)`

= `(1 + tan(pi/4 - (2x)/3))/(1 - tan(pi/4 - (2x)/3))                                ...["Dividing by" cos(pi/4 - (2x)/3)` 

= `(tan  pi/4 + tan(pi/4 - (2x)/3))/(1 - tan  pi/4. tan(pi/4 - (2x)/3))                        ...[∵ tan  pi/4 = 1]`

= `tan[pi/4 + pi/4 - (2x)/3]`

= `tan(pi/2 - (2x)/3)`

= `cot((2x)/3)`

∴ y = `cot^-1[cot((2x)/3)] = (2x)/(3)`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"((2x)/3)`

= `(2)/(3)"d"/"dx"(x)`

= `(2)/(3) xx 1`

= `(2)/(3)`

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

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.2 | Q 7.12 | पृष्ठ ३०

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