Question

Find the derivative of $|x|+a_0{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+....+a_{n-1}x+a_n$

• A. $\frac{x}{|x|}+n{a}_0{x}^{n-1}+(n-1)a_1{x}^{n-2}+(n-2)a_2{x}^{n-3}+....+a_{n-1}$
• B. $1+n{a}_0{x}^{n-1}+(n-1)a_1{x}^{n-2}+(n-2)a_2{x}^{n-3}+....+a_{n-1}$
• C. $\frac{x}{|x|}+n{a}_0{x}^{n-1}+(n-1)a_1{x}^{n-2}+(n-2)a_2{x}^{n-3}+....+a_n$
• D. None of these

Answer 1

The correct option is A $\frac{x}{|x|}+n{a}_0{x}^{n-1}+(n-1)a_1{x}^{n-2}+(n-2)a_2{x}^{n-3}+....+a_{n-1}$

We can write $|x|$ as $\sqrt{x^2}$
so, $\frac{d\left(\sqrt{x^2}\right)}{dx}=\frac{2x}{2\sqrt{x^2}}=\frac{x}{|x|}$
and, $\frac{d\left(a_0{x}^n\right)}{dx}=n{a}_0{x}^{n-1}$
Similarly, $\frac{d\left(a_{n-1}x\right)}{dx}=a_{n-1}$ and $\frac{d\left(a_n\right)}{dx}=0$
Hence, option A is the correct option.