How do you find the first and second derivative of x^2 lnxx2lnx?

1 Answer
ali ergin
Aug 31, 2016

(d y)/(d x)=2x*l n(x)+x" 'The first derivative '"dydx=2x⋅ln(x)+x 'The first derivative '
(d^2y)/(d x^2)=2*l n(x)+3" 'The second derivative '"d2ydx2=2⋅ln(x)+3 'The second derivative '

Explanation:

y=x^2* l n (x)y=x2⋅ln(x)

u=x^2" ; " u'=2x

v=l n (x)" ; "v^'=1/x

y=u*v

(d y)/(d x)=u^'*v+v^'*u

(d y)/(d x)=2x* l n(x)+1/cancel(x)*cancel(x)^2

(d y)/(d x)=2x*l n(x)+x" 'The first derivative '"

a=2x" ; "a^'=2
b=l n(x)" ; "b^'=1/x

(d^2y)/(d x^2)=a^'*b+b^'*a+1

(d^2y)/(d x^2)=2*l n(x)+1/cancel(x)*2cancel(x)+1

(d^2y)/(d x^2)=2*l n(x)+2+1

(d^2y)/(d x^2)=2*l n(x)+3" 'The second derivative '"

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