Question

Find the derivative of the function at the given number.

.$f\left(x\right)=1-3x^2,at2$

Short answer

The derivative of the function at2 is.$-12$

Step-by-step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=1-3x^2,at2$

Step 2. Formula used.

The derivative of a function f at a number,denoted by$f\text{'}\left(a\right),$is

.$f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$

if this limit exists.

Step 3. Finding the slope of the tangent line at a given point.

According to the definition of a derivative, with$a=2$, we have

$\begin{array}{c}f\text{'}\left(2\right)=\underset{h\to 0}{\lim }\frac{f(2+h)-f\left(2\right)}{h}Definitionoff\text{'}\left(2\right)\\ =\underset{h\to 0}{\lim }\frac{[1-3{(2+h)}^2]-[1-3{\left(2\right)}^2]}{h}f\left(x\right)=1-3x^2\\ =\underset{h\to 0}{\lim }\frac{[1-3(2^2+h^2+4h\left)\right]+11}{h}Expand\\ =\underset{h\to 0}{\lim }\frac{[1-12-3h^2-12h]+11}{h}\\ =\underset{h\to 0}{\lim }\frac{-3h^2-12h}{h}Simplify\\ =\underset{h\to 0}{\lim }(-3h-12)Cancelh\\ =-12\end{array}$

Therefore, the derivative of the function at 2is$-12$ .