The following is a problem from chapter 11, "Significance of the Derivative", from Spivak's Calculus
- Redo problem 10-18(c) when
$$(f')^2=f-\frac{1}{f^2}\tag{1}$$
Here is problem 10-18(c)
10-18. (c) Suppose that the function $f>0$ has the property that
$$(f')^2=f+\frac{1}{f^3}\tag{2}$$
Find a formula for $f''$ in terms of $f$.
10-18 is relatively straightforward to solve.
First, note that if $f'(x)=f(x)+\frac{1}{f(x)^3}=0$ then $f(x)^4=-1$, which never occurs for any $x$. Hence $f' \neq 0$.
Just take the derivative of both sides of $(2)$
$$2f'f''=f'-3f^{-4}f'$$
Therefore $f''(x)=\frac{1}{2}-\frac{3}{2f^{(4)}(x)}$.
Let's consider the solution to 59.
First, note that if $f'(x)=0$ then $0=f(x)-\frac{1}{f^2(x)} \implies f(x)=1$
Differentiate both sides of $(1)$
$2f'(x)f''(x)=f'(x)+\frac{2f'(x)}{f^3(x)}$
Dividing by $f'(x)$, we have for $f(x) \neq 1$ that
$$f(x)''=\frac{1}{2}+\frac{1}{f(x)^3}\tag{3}$$
At this point, according to the solution manual we can show that $(3)$ also holds if $f(x) = 1$, by applying the following theorem
Theorem 7: Suppose $f$ is continuous at $a$, $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim\limits_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and
$$f'(a)=\lim\limits_{x \to a} f'(x)$$
How do we show that $(3)$ also holds for $f(x)=1$ using this theorem?