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Find \(\cfrac{dy}{d\text x}\), when

y = (sin x)cos x + (cos x)sin x

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Best answer

let y = (sin x)cos x + (cos x)sin x

⇒ y = a + b

where a= (sin x)cos x ; b = (cos x)sin x

  \(\Bigg\{\) Using chain rule, \(\cfrac{d(u +a)}{d\text x}=\cfrac{du}{d\text x}+\cfrac{da}{d\text x}\) where a and u are any variables \(\Bigg\}\)

a= (sin x)cos x

Taking log both the sides:

⇒ log a= log (sin x)cos x

⇒ log a= cos x log (sin x)

{log xa = a log x}

Differentiating with respect to x:

b = (cos x)sin x

Taking log both the sides:

⇒ log b= log (cos x)sin x

⇒ log b= sin x log (cos x)

{log xa = a log x}

Differentiating with respect to x:

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