Let y = xx + (sin x)x
⇒ y = a + b
where a = xx ; b = (sin x)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= xx
Taking log both the sides:
⇒ log a= log (x)x
⇒ log a= x log x {log xa = a log x}
Differentiating with respect to x:
b = (sin x)x
Taking log both the sides:
⇒ log b= log (sin x)x
⇒ log b= x log (sin x)
{log xa = a log x}
Differentiating with respect to x: