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find dy/dx wheny=(sin x)^x + sin^-1(sqrt x)
Let y1=(sinx)sinxtaking log on both sides, getlogy1=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√xdiff.w.r.to x , we getdy2/dx=(1/√(1-x)).(1/2√x)=(1/2).√x(1-x)Hence dy/dx=dy1/dx+dy2/dxdy/dx =(sinx)sinx.cosx[1+log(sinx)]+1/[2√{x(1-x)}] Ans.Thanks & RegardsRinkoo GuptaAskIITians Faculty
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
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
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