Question
If y=ln(xa+bx)x,then x3d2ydx2 is equal to
Answer: Option D
:
D
∵y=ln(xa+bx)x=x(lnx−ln(a+bx))
or(yx)=lnx−ln(a+bx)
Differentiating both sides w.r.t.x,then
xdydx−y.1x2=1x−ba+bx=ax(a+bx)⋯(i)
or(xdydx−y)=(axa+bx)
Again taking logarithm on both sides, then
ln(xdydx−y)=ln(ax)−ln(a+bx)
Differetiating both sides w.r.t.x, then
xd2ydx2+dydx−dydx(xdydx−y)=1x−ba+bx=ax(a+bx)=(xdydx−y)x2[FromEq.(i)]orx3d2ydx2=(xdydx−y)2
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:
D
∵y=ln(xa+bx)x=x(lnx−ln(a+bx))
or(yx)=lnx−ln(a+bx)
Differentiating both sides w.r.t.x,then
xdydx−y.1x2=1x−ba+bx=ax(a+bx)⋯(i)
or(xdydx−y)=(axa+bx)
Again taking logarithm on both sides, then
ln(xdydx−y)=ln(ax)−ln(a+bx)
Differetiating both sides w.r.t.x, then
xd2ydx2+dydx−dydx(xdydx−y)=1x−ba+bx=ax(a+bx)=(xdydx−y)x2[FromEq.(i)]orx3d2ydx2=(xdydx−y)2
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