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Differentiate the following:y = sin3x + cos3x

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प्रश्न

Differentiate the following:
y = sin3x + cos3x

योग
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उत्तर

y = sin3x + cos3x

Here u = sin3x = (sin x)3 

⇒ `("d"u)/("d"x)` = 3(sin x)2(cos x)

= 3sin2x cosx

v = cos3x = (cos x)

⇒ `("d"u)/("d"x)` = 3(cos x)2 (– sin x)

= – 3 sin x cos2x

Now y = u + v

⇒ `("d"y)/("d"x) = ("d"u)/("d"x) + ("d"v)/("d"x)`

= 3 sin2x cos x – 3sin x cos2x

= 3 sin x cos x (sin x – cos x)

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Differentiation Rules
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Differential Calculus - Differentiability and Methods of Differentiation - Exercise 10.3 [पृष्ठ १६४]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 10 Differential Calculus - Differentiability and Methods of Differentiation
Exercise 10.3 | Q 22 | पृष्ठ १६४

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