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Grade: 12
        
find dy/dx when
y=(sin x)^x + sin^-1(sqrt x)
5 years ago

Answers : (2)

Rinkoo Gupta
askIITians Faculty
80 Points
							

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

5 years ago
Yash Yarav
13 Points
							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
						
one year ago
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