Mistake in taking second differential

Question

In my book exercise I need to take total differential of first and second order from function $$ u=f(t), t = xyz $$ with first order all is good, I get it like that $$ du = \frac{du}{dt}dt = f'(t)dt $$ $$ dt = yzdx +xzdy+xydz$$ $$ du=f'(t)*(yzdx+xzdy+xydz)$$ but second order does not matchup with answer in the end of the book. I do like that $$ d^2u = \frac{d^2u}{dt^2}dt^2 = f''(t)dt^2 $$ $$ dt^2= (yzdx +xzdy+xydz)^2$$ $$ d^2u=f''(t)*(yzdx +xzdy+xydz)^2 $$ Answer from book is $$ d^2u=f''(t)*(yzdx +xzdy+xydz)^2 +2f'(t)*(zdxdy+ydxdz+xdydz) $$ Why does there appear first derivative?

Answer 1

Observe that $$ d^2u=d(du)=d(f'(t)dt)=df'(t)dt+f'(t)d(dt)=f''(t)dt^2+f'(t)d^2t. $$ Basically, be very careful using tricks to 'cancel' second derivatives or differentials. See Cancellation rules for partial derivatives for more details.

If you want further justification, see what happens if you try out your formula in a simple univariate case, like $u=t^2$ and $t=x^3$.