For the first derivative, use the chain rule.
y = ln(u)y=ln(u) and u = x^2u=x2. 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!
