Let y = x(sin x - cos x) + \(\cfrac{\text x^2-1}{\text x^2+1}\)
⇒ y = a + b
where a = x(sin x - cos x); b = \(\cfrac{\text x^2-1}{\text x^2+1}\)
{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}
a = x(sin x - cos x)
Taking log both the sides:
⇒ log a = log x(sin x - cos x)
⇒ log a = (sin x - cos x) log x
{log xa = a log x}
Differentiating with respect to x:
{\(\cfrac{d(log\,u)}{d\text x}=\cfrac{1}{u}\cfrac{du}{d\text x};\) 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}
{Using chain rule, \(\cfrac{d(u+a)}{d\text x}=\cfrac{du}{d\text x}\) where a is any constant and u is any variable}