Not a dumb question nor is there a nice answer though you can do the differentiation with simple differentiation rules.

If h(x)= f(g(x)) then, by the chain rule, h'= f'(g(x))g'(x). Differentiate again using the chain rule and the product rule: h''= f''(g(x))g'(x)g'(x)+ f'(g(x))g''(x)= f''(x)g(x)(g'(x))^2+ f'(g(x))g''(x). Once more, h'''= f'''(g(x))(g'(x))^3+ 3f''(x)g'(x)^2g''(x)+ f'(x)g'''(x).

You may have been thinking of the higher derivatives of a product which has a nicer formula:

(fg)''= f''g+ 2f'g'+ fg'', (fg)'''= f'''g+ 3f''g'+ 3f'g''+ g'''.

Generally,