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Differentiate w.r.t. x the function: cos-1 x22x+7,-2<x<2 - Mathematics

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प्रश्न

Differentiate w.r.t. x the function:

`(cos^(-1)  x/2)/sqrt(2x+7), -2 < x < 2`

योग

उत्तर

Let, y = `(cos^-1 x/2)/(sqrt(2x + 7)) = u/v`

`therefore u = cos^-1  x/2, v = sqrt(2x + 7)`

Now,  u = `cos^-1  x/2`

On differentiating with respect to x,

`(du)/dx = d/dx cos^-1  x/2`

`= - 1/(sqrt(1 - x^2/4)) d/dx (x/2)`

`= - 2/(sqrt(4 - x^2)) * 1/2`

`= (-1)/sqrt(4 - x^2)`                   ...(1)

and  v = `sqrt(2x + 7)`

On differentiating with respect to x,

`(dv)/dx = 1/2 (2x - 7)^(1/2 - 1) d/dx (2x - 7)`

`= 1/2 (2x - 7)^(- 1//2) (2) = 1/(sqrt(2x + 7))`            ...(2)

y = `u/v`

∴ `dy/dx = (v  (du)/dx - u  (dv)/dx)/v^2`      ... [(1) and (2) substituting the value of]

`= (- 1/(sqrt(4 - x^2)) xx sqrt(2x + 7) - (cos^-1  x/2)/sqrt(2x + 7))/((2x + 7))`

`= - [1/(sqrt(4 - x^2) sqrt(2x + 7)) + (cos^-1  x/2)/(2x + 7)^(3//2)]`

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अध्याय 5: Continuity and Differentiability - Exercise 5.9 [पृष्ठ १९१]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.9 | Q 5 | पृष्ठ १९१

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