How can I find the derivative.

Question

How can I find the derivative of $$f(x) =1-3x^2$$

using the alternative definition?

Help would be appreciated.

Answer 1

Using the power rule, $f'(x) = -6x$.

Using the definition of the derivative, \begin{align*} f'(x) &= \lim_{t\to x} \frac{f(t) - f(x)}{t-x} \\ &= \lim_{t\to x} \frac{1-3t^2 - (1-3x^2)}{t-x} \\ &= \lim_{t\to x} \frac{-3t^2 + 3x^2}{t-x} \\ &= \lim_{t\to x} \frac{3(x^2-t^2)}{t-x} \\ &= \lim_{t\to x} \frac{3(x-t)(x+t)}{t-x} \\ &= \lim_{t\to x} -3(x+t) \\ &= -3\cdot 2x = -6x \end{align*} which agrees with our first solution.

Answer 2

$f’(x)=\displaystyle \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$

$\implies f’(x)=\displaystyle \lim_{h\to 0} \dfrac{1–3(x+h)^2–1+3x^2}{h}$

$\implies f’(x)=\displaystyle \lim_{h\to 0} \dfrac{-6xh-3h^2}{h}$

$\implies f’(x)=\displaystyle \lim_{h\to 0} -6x-3h$

$\implies f’(x)=-6x $

Answer 3

Applying linearity of the derivative, the derivative of a constant being $0$ and the derivative of $x^2$ being $2x$ we get

$$f'(x)=(1)'-3(x^2)'=-6x$$

Now let's have a look at the definition

$$\begin{align} f'(x)=&\lim_{h\to 0}{f(x+h)-f(x)\over h}\\ =&\lim_{h\to 0}{1-3(x+h)^2-(1-3x^2)\over h}\\ =&\lim_{h\to 0}{1-3x^2-6hx-h^2-1-3x^2\over h}\\ =&\lim_{h\to 0}{-6hx-h^2\over h}\\ =&\lim_{h\to 0}{-6x-h}=-6x \end{align}$$

It is consistent.