find dy/dx when

y=x^(cos x) + (sin x)^ tan x

Let y1=x^cosx

^{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)]

=x^cosx[(cosx)/x –sinx .logx]

let y2=sinx^tanx

^{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=sinx^tanx[1+sec²x . logsinx]

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

Thanks & Regards

Rinkoo Gupta

askIITians Faculty