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Differentiate the function with respect to x. cos x3 . sin2 (x5) - Mathematics

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प्रश्न

Differentiate the function with respect to x. 

cos x3 . sin2 (x5)

योग
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उत्तर

Let, y = cos x3 . sin2 (x5)

`dy/dx = d/dx [cos x^3. sin^2 (x^5)]` 

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

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

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

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

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

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.2 | Q 6 | पृष्ठ १६६

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