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

when y = (sin x)x + sin-1√x

1 Answer

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Best answer

Let y = (sin x)x + sin-1√x

⇒ y = a + b

where, a = (sin x)x; b = sin-1√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)x

Taking log both the sides:

⇒ log a= log (sin x)x

⇒ log a= x log (sin x)

{log xa = a log x}

Differentiating with respect to x:

Put the value of a= (sin x)x :

⇒ \(\cfrac{da}{d\text x}\) = (sin x)x{x cot x + log(sin x)}

b = sin-1√x

⇒ b = sin-1(x)\(\frac12\) 

Differentiating with respect to x:

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