Question

What is the second derivative of #f(x)=sin(1/x^2) #?

Answer 1

`f'(x)=cos(1/x)*-1/x^2`
`f=cos(1/x), g= -1/x^2-> f'=-sin(1/x)(-1/x^2), g'=2/x^3`
`f''(x)=fg'+gf'=cos(1/x)(2/x^3)+(-1/x^2)sin(1/x)(1/x^2)=2/x^3cos(1/x^2)+(-1/x^4)sin(1/x)`

Explanation:

Find the first derivative using chain rule. To find the second derivative use the product rule. So separate the functions into f and g then find f' and g' and then apply the product rule.