find dy/dx when

y=(sin x)^x + sin^-1(sqrt x)

Let y₁=(sinx)^sinx

taking log on both sides, get

logy1=sinx . log (sinx)

diff.w.r.to x , we get

(1/y₁).dy₁/dx=sinx.(1/sinx).cosx+log(sinx).cosx

=cosx+cosx.log(sinx)

dy1/dx=(sinx)^sinx[cosx+cosx.log(sinx)]

=(sinx)^sinx.cosx[1+log(sinx)]

let y₂=sin^-1√x

diff.w.r.to x , we get

dy₂/dx=(1/√(1-x)).(1/2√x)

=(1/2).√x(1-x)

Hence dy/dx=dy₁/dx+dy₂/dx

dy/dx =(sinx)^sinx.cosx[1+log(sinx)]+1/[2√{x(1-x)}] Ans.

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty

let, u=sinx^x taking log on both side log u= logsinx^x log u= xlogsinx diff. w.r.t X (1/u).du/dx = x.(1/sinx).cosx + logsinx du/dx = u[x.(cosx/sin) + logsinx] du/dx = sinx^x[xcotx + logsinx] Now,let v = sin^-1√x diff. w.r.t X dv/dx=1/√1-x.1/2√x dv/dx = 1/2√x.√1-x dy/dx = du/dx+dv/dx dy/dx = sinx^x[xcotx + logsinx]+1/2√x.√1-x thanks Rao Yash Yadav