Find the second derivative y= x^2lnx

1 answer:

3 0

First we use product rule
y=x^2lnx
dy/dx = x^2 d/dx (lnx) + lnx d/dx (x^2)
dy/dx = x^2 (1/x) + lnx (2x)
dy/dx = x + 2xlnx

now taking second derivative:
d2y/dx2 = 1 + 2[x (1/x) + lnx (1)]
d2y/dx2 = 1 + 2[1+lnx]
1+2+2lnx
3+2lnx is the answer

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