Question

Use the differentiation rules developed in this section to find

the derivatives of the functions

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

Short answer

The derivatives of the given function is$2x+2-9x^2$

Step-by-step solution

Step1. Given information 

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

We need to find out the derivatives of the given function.

Step2. finding the derivatives of the function

the given function is $f\left(x\right)=x^2+x(2-3x^2)$. When opening the parenthesis in the given function multiplying x , we get

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

Now we can find out the derivatives of this expansion form then,

localid="1649266685938" role="math" $$\begin{gathered} \frac{d}{d\mathrm{x}}({\mathrm{x}}^2+2\mathrm{x}-3{\mathrm{x}}^3)=\frac{d\left({\mathrm{x}}^2\right)}{d\mathrm{x}}+\frac{d\left(2\mathrm{x}\right)}{d\mathrm{x}}-\frac{d\left(3{\mathrm{x}}^3\right)}{d\mathrm{x}} \\ \mathrm{derivative}\mathrm{of}\mathrm{an}\mathrm{identity}\mathrm{function}\mathrm{x},\frac{d\left(\mathrm{x}\right)}{d\mathrm{x}}=1\mathrm{so}\frac{d\left(2\mathrm{x}\right)}{d\mathrm{x}}=2\frac{d\left(\mathrm{x}\right)}{d\mathrm{x}} \\ =2 \\ \mathrm{The}\mathrm{derivative}\mathrm{of}\frac{d}{d\mathrm{x}}\left({\mathrm{x}}^{\mathrm{n}}\right)={\mathrm{nx}}^{\mathrm{n}-1},\mathrm{for}\mathrm{any}\mathrm{non}\mathrm{zero}\mathrm{rational}\mathrm{number}\mathrm{n}\mathrm{using}\mathrm{power}\mathrm{rule},\mathrm{then} \\ \frac{d\left({\mathrm{x}}^2\right)}{d\mathrm{x}}=2\mathrm{x} \\ \frac{d\left(3{\mathrm{x}}^3\right)}{d\mathrm{x}}=3\frac{d\left({\mathrm{x}}^3\right)}{d\mathrm{x}} \\ =3\times 3{\mathrm{x}}^2 \\ =9{\mathrm{x}}^2 \\ \mathrm{thus}\mathrm{the}\mathrm{derivative}\mathrm{of}\mathrm{given}\mathrm{function},\frac{d}{d\mathrm{x}}({\mathrm{x}}^2+\mathrm{x}(2-3{\mathrm{x}}^3)=2\mathrm{x}+2-9{\mathrm{x}}^2 \end{gathered}$$