Question

Use the differentiation rules developed in this section to find

the derivatives of the functions

$f\left(x\right)=2x-3+4x^2$

Short answer

The derivative of$\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}2+8\mathrm{x}$

Step-by-step solution

Step1. Given information

Here function is $f\left(x\right)=2x-3+4x^2$

We need to find out the the derivatives of the functions

Step2. finding the derivatives of the function

When we differentiate the given function, by using differentiation rules we get

$$\begin{gathered} \frac{d}{d\mathrm{x}}(2\mathrm{x}-3+4{\mathrm{x}}^2)=\frac{d\left(2\mathrm{x}\right)}{d\mathrm{x}}-\frac{d\left(3\right)}{d\mathrm{x}}+\frac{d\left(4{\mathrm{x}}^2\right)}{d\mathrm{x}} \\ \mathrm{since}\mathrm{derivative}\mathrm{of}\mathrm{a}\mathrm{constant}\mathrm{is}0,\mathrm{so}\frac{d\left(3\right)}{d\mathrm{x}}=0\mathrm{and} \\ \mathrm{derivative}\mathrm{of}\frac{d}{d\mathrm{x}}\left({\mathrm{x}}^{\mathrm{n}}\right)={\mathrm{nx}}^{\mathrm{n}-1},\mathrm{for}\mathrm{any}\mathrm{non}\mathrm{zero}\mathrm{rational}\mathrm{number}\mathrm{n}\mathrm{using}\mathrm{power}\mathrm{rule}\mathrm{then}\frac{d\left(4{\mathrm{x}}^2\right)}{d\mathrm{x}}=4\frac{d\left({\mathrm{x}}^2\right)}{d\mathrm{x}} \\ =4\times 2\mathrm{x} \\ =8\mathrm{x} \\ \mathrm{we}\mathrm{know}\mathrm{that}\mathrm{derivetive}\mathrm{of}\mathrm{identity}\frac{d\left(\mathrm{x}\right)}{d\mathrm{x}}=1,\mathrm{so}\frac{d\left(2\mathrm{x}\right)}{d\mathrm{x}}=2\frac{d\left(\mathrm{x}\right)}{d\mathrm{x}} \\ =2 \\ \mathrm{thus}\mathrm{the}\mathrm{derivative}\mathrm{of}\mathrm{given}\mathrm{function},\frac{d}{d\mathrm{x}}(2\mathrm{x}-3+4{\mathrm{x}}^2)=2-0+8\mathrm{x} \\ =2-8\mathrm{x} \end{gathered}$$