Chapter 10.2, Problem 9QR

To determine

To find the derivative of the given function.

Answer to Problem 9QR

  $\frac{dy}{dx}=\frac{x^{2n-1}}{\left(2n-1\right)!}$

Explanation of Solution

Given:

  $y=\frac{{\left(x+a\right)}^{2n}}{\left(2n\right)!}$

Calculation:

  $y=\frac{{\left(x+a\right)}^{2n}}{\left(2n\right)!}$

Differentiating with respect to x, we get

  $\begin{array}{l}\frac{dy}{dx}=\frac{1}{\left(2n\right)!}\cdot \left(2n\right)x^{2n-1}\\ =\frac{\left(2n\right)x^{2n-1}}{\left(2n\right)\cdot \left(2n-1\right)!}\\ =\frac{x^{2n-1}}{\left(2n-1\right)!}\end{array}$