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: