Abstract Differentiation

Problem

Secant line and tangent line on a curve.

Consider a function that satisfies $f(x+y) = f(x) + f(y) + x^2y + xy^2$ for $x, y \in \R$, and follows the property

$$\lim_{x\rightarrow0} \frac{f(x)}{x} = 1.$$

Determine $f'(x)$.

Hint

(1) Apply the limit definition of the derivative: $f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$.

Solution

Rearrange $f(x+y) = f(x) + f(y) + x^2y + xy^2$ into $f(x+y) - f(x) = f(y) + x^2y + xy^2$.

Then divide by $y$ on both sides $\frac{f(x+y) - f(x)}{y} = \frac{f(y)}{y} + x^2 + xy$.

Consider the definition of the derivative

$$f'(x) = \lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}.$$

Apply this limit to the rearranged expression (with $h$ replaced with $y$):

$$\lim_{y\rightarrow0} \frac{f(x+y) - f(x)}{y} = \lim_{y\rightarrow0}\left[\frac{f(y)}{y} + x^2 + xy \right].$$

Since $\lim_{x\rightarrow0} \frac{f(x)}{x} = 1$, it follows that $\lim_{y\rightarrow0} \frac{f(y)}{y} = 1$; thus,

$$f'(x) = \lim_{y\rightarrow0} \frac{f(x+y) - f(x)}{y} = 1 + x^2.$$