Why do we not need to consider the cases x=0 or x=1 ?

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We have seen in the previous page how the derivative is defined: For a function f(x), its derivative at x=a is defined by

$\begin{displaymath}f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a} = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}\cdot\end{displaymath}$

Let us give some examples.

Example 1. Let us start with the function f(x) = x². We have

$\begin{displaymath}\frac{f(a + h) - f(a)}{h} = \frac{(a + h)^2 - a^2}{h} = \frac{2 a h + h^2}{h} = 2a + h.\end{displaymath}$

$\begin{displaymath}\lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h} = 2 a.\end{displaymath}$

which means f '(a) = 2a.

What about the derivative of f(x) = xⁿ. Similar calculations, using the binomial expansion for (x+y)ⁿ (Pascal's Triangle), yield

$\begin{displaymath}f'(a) = n a^{n-1}\cdot\end{displaymath}$

Example 2. Consider the function f(x)=1/x for $x\not=0$ . We have

$\begin{displaymath}\frac{f(a + h) - f(a)}{h} =\frac{\frac{1}{a+h}-\frac{1}{a}}{h}=\frac{a-(a+h)}{h a (a+h)}=\frac{-h}{h a (a+h)}\end{displaymath}$

Consequently, $\begin{displaymath}\lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h} =\lim_{h\rightarrow 0} \frac{-1}{a (a+h)}=-\frac{1}{a^2}.\end{displaymath}$

Have you noticed? The algebraic trick in both of the examples above has been to factor out "h" in the numerator, so that we can cancel it with the "h" in the denominator! This is what you try to do whenever you are asked to compute a derivative using the limit definition.

I hope this helps. :)

answered Dec 18, 2015 by Mathical Level 10 User (57.4k points)

Why do we not need to consider the cases x=0 or x=1 ?

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