Question

Find the derivative of the function at the given number.

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

Short answer

The derivative of the function at $-1$is7.

Step-by-step solution

Step 1. Given information.

The function here given is,

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

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,$a=-1$ we have

$\begin{array}{c}f\text{'}(-1)=\underset{h\to 0}{\lim }\frac{f(-1+h)-f(-1)}{h}Definitionoff\text{'}(-1)\\ =\underset{h\to 0}{\lim }\frac{\left[\right(-1+h)-3{(-1+h)}^2]-[-1-3{(-1)}^2]}{h}f\left(x\right)=x-3x^2\\ =\underset{h\to 0}{\lim }\frac{[-1+h-3(1+h^2-2h\left)\right]+4}{h}Expand\\ =\underset{h\to 0}{\lim }\frac{-1+h-3-3h^2+6h+4}{h}\\ =\underset{h\to 0}{\lim }\frac{-3h^2+7h}{h}Simplify\\ =\underset{h\to 0}{\lim }(-3h+7)Cancelh\\ =7\end{array}$

Therefore, the derivative of the function at $-1$is7 .