For the first derivative, use the chain rule.
#y = ln(u)# and #u = x^2#. Then #y' = 1/u# and #u' = 2x#.
#y' = 2x xx 1/u = 2x xx 1/(x^2) = (2x)/x^2#
So, the first derivative is #(2x)/x^2#. The second derivative can be determined by differentiating the first.
By the quotient rule:
Let #y = (g(x))/(h(x))#, so that #g(x) = 2x# and #h(x) = x^2#. The derivative is given by #y' = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2#
The derivative of #g(x)# is #2# and the derivative of #h(x)# is #2x#.
We can now substitute into the formula and calculate. Note: the notation #y''# is used to show that we're finding the second derivative, and not the first, as would the notation #y'#. Similarly, #y'''# would signify the third derivative.
#y'' = (2 xx x^2 -2x xx 2x)/(x^2)^2#
#y'' = (2x^2 - 4x^2)/x^4#
#y'' = (-2x^2)/x^4#
#y'' = -2/x^2#
Hence, the second derivative is #y'' = -2/x^2#.
Hopefully this helps!
