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Differentiate the function with respect to x. cosx3.sin2(x5) - Mathematics

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प्रश्न

Differentiate the function with respect to x. 

`cos x^3. sin^2 (x^5)`

योग

उत्तर

`cos x^3 . sin^2 (x^5)`

Let y = `cos x^3 . sin^2 (x^5)`

`dy/dx = (d(cos x^3. sin^2 (x^5)))/dx` by quotient rule

`= (cos x^3  d sin^2 (x^5))/dx + (sin^2 (x^5) d cos x^3)/dx`  By the chain rule

`= (cos x^3 (2 sin (x^5)). d sin (x^5))/dx + (sin^2 (x^5) (- sin x^3). d  x^3)/dx`

`= (cos x^3 (2 sin (x^5)). cos x^5 dx^5)/dx + sin^2 (x^5) (- sin x^3) 3x^2` 

`= cos x^3 2 sin (x^5) cos x^5 . 5x^4 + sin^2 (x^5)(- sin x^3) 3x^2`

`= 10  x^4 sin x^5 cos x^5 cos x^3 - sin x^3 sin^2 sin^2 x^5`

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अध्याय 5: Continuity and Differentiability - Exercise 5.2 [पृष्ठ १६६]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.2 | Q 6 | पृष्ठ १६६

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