Question

Find the derivative of the function at the given number.

.$f\left(x\right)=x+x^3,at1$

Short answer

The derivative of the function at1 is.4

Step-by-step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=x+x^3,at1$

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

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

Therefore, the derivative of the function at 1is 4.