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Grade: 12
        
find dy/dx when
y=x^(cos x) + (sin x)^ tan x
5 years ago

Answers : (1)

Rinkoo Gupta
askIITians Faculty
80 Points
							

Let y1=xcosx

taking log on boths ides, we get

logy1=cosx.logx

now diff.w.r.to x , we get

(1/y1)dy1/dx=cosx.(1/x)+(logx).(-sinx)

dy1/dx=y1[(cosx)/x-sinx.(logx)]

=xcosx[(cosx)/x –sinx .logx]

let y2=sinxtanx

taking log on both sides, we get

logy2=tanxlogsinx

diff. w.r. to x , get

(1/y2)dy2/dx=tanx.(1/sinx).cosx+logsinx.sec2x

dy2/dx=sinxtanx[1+sec2x . logsinx]

Hence dy/dx = xcosx[(cosx)/x –sinx .logx]+ sinxtanx[1+sec2x . logsinx]

Thanks & Regards

Rinkoo Gupta

askIITians Faculty

5 years ago
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