Let and
be twice-differentiable functions such that
- If
, compute
.
- If
, compute
.
- Compute
.
- Using the rule for the derivative of a quotient we have,
Substituting the given expressions for
, and
we have,
- Using the product rule (more precisely, the product rule for more than two terms that we proved in this exercise) we have,
Then, we evaluate at 0,
since we are given
.
- Since we are given that
and
are twice differentiable functions, we know that
and
are differentiable; hence,
and
are continuous since differentiability implies continuity (from Section 4.4 of Apostol). From the definition of continuity of a function at a point we know this means,
Furthermore,
since
and
. So we can then compute,
since we are given
.