y=ln(x×ex2)
Let u=x×ex2⟹y=ln(u)⟹dudy=u1
Let f=x=x21⟹f′=21x21−1=21x−21
Let g=ex2⟹g′=2xex2
We then use product rule of differentiation to differentiate u=x×ex2 with respect to x
dxdu=f′g+fg′
=21x−21ex2+x(2xex2)
We then use chain rule of differentiation to differentiate y=ln(x×ex2) with respect to x
dxdy=dudy×dxdu
=u1×(21x−21ex2+x(2xex2))
=x×ex21×(21x−21ex2+x(2xex2))
=x×ex21×(2(x)1ex2+x(2xex2))
=2xex2ex2+x(ex2)x(2xex2
=2x1+2x
∴dxdy=2x1+4x2
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