11 Find x31nx dx4 Solution11 ddxx3lnx x31xlnx3x2 x23x2lnx x

Solution

11

d/dx(x^3lnx)

= x^3(1/x)+lnx(3x^2)

=x^2+3x^2(lnx)

=x^2(1+3lnx)

Differentiating

=x^2(3/x)+(1+3lnx)(2x)

=3x+(1+3lnx)2x

=3x+2x+6xlnx

=5x+6(xlnx)

Differentiating

=5+6(x(1/x)+lnx(1))

=5+6(1+lnx).

Differentiating

=0+6(0+1/x).

=6/x.

14.

Given that

xy^4-x^2y=x+3y

Differentiating on both sides

x(4y^3dy/dx)+y^4(1)-{x^2(dy/dx)+y(2x)}=1+3dy/dx.

dy/dx(4y^3x-x^2)+y^4-2xy=1+3.dy/dx

dy/dx(4xy^3-x^2-3)=2xy+1-y^4.

So dy/dx= (2xy+1-y^4)/(4xy^3-x^2-3).