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: