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” dy .= “"fnnddx--—'+(sinx) oL |
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Answer» y= x^sinx + (sinx)^cosxlet x^sinx = u and (sinx)^cosx= vdy/dx = du/dx + dv/dxu= x^sinx+++Taking log on both sideslog u = log(x^sinx) = sinx log xDifferntiating both sides1/u du/dx= cosx logx + sinx/x (product rule)du/dx= u {cosxlogx+sinx/x}du/dx= x^sinx(cosxlogx+sinx/x) Now, v = (sinx)^cosxTaking log on both sideslog v = log {(sinx)^cosx} = cosxlog sinxDifferentiating1/v. dv/dx = (-sinx)log sinx+ cos x (cos x/ sin x)dv/dx = v { cos x cot x - sinx log sinx}dv/dx = (sinx)^cos x (cos x cot x - sinx log sinx)=> dy/dx = {x^sinx(cosxlogx+sinx/x)} + {(sinx)^cosx (cosx cot x -sinxlogsinx)}Therefore, dy/dx = x^sinx (cosxlog x + sinx/x) + sinx^cosx (cosx cotx - sinx log sinx) |
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