Q.40 - Calculus

Question

Use the differentiation rules developed in this section to find

the derivatives of the functions

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

Step-by-Step Solution

Answer

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

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,

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