y = $(sinx)^x+sin^{-1}√x$ Let u = $(sinx)^x$ and v = $sin^{-1}√x$ Now y = u + v ${dy}/{dx}$ = ${du}/{dx}+{dv}/{dx}$ ... (i) Consider u = $(sinx)^x$ Taking logarithms on both the sides, we have, logu = xlog (sin x) Differentiating with respect to x, we have, $1/u . {du}/{dx}$ = log (sin x) + $x/{sinx}$ . cos x ⇒ ${du}/{dx}$ = $(sinx)^x$ (log (sin x) + x cot x) ... (ii) Consider v = $sin^{-1}√x$ ${dv}/{dx}$ = $1/√{1-x}$ × $1/{2√x}$ ... (iii) From (i), (ii) and (iii) We get , ${dy}/{dx}$ = $(sinx)^x$ (log (sinx) + x cot x) + $1/{2√x√{1-x}}$