Chapter 7.3, Problem 1QR

To determine

  $\text{ To find }\frac{dy}{dx}\text{.}$

Answer to Problem 1QR

  $\text{The required solution is, }\frac{dy}{dx}=3x^2\sin 2x+2x^3\cos 2x.$

Explanation of Solution

Given information: $\text{The equation is},y=x^3\sin 2x$

Calculation : $\frac{dy}{dx}\text{ is obtained as,}$

  $\begin{array}{l}\text{Find the derivative by using product rule, }\frac{d}{dx}\left(uv\right)=u\text{'}v+uv\text{'}\\ \\ y=x^3\sin 2x\\ \frac{dy}{dx}=3x^2\sin 2x+x^3\left(2\cos 2x\right)\\ =3x^2\sin 2x+2x^3\cos 2x\\ \text{Hence, the required solution is, }\frac{dy}{dx}=3x^2\sin 2x+2x^3\cos 2x.\end{array}$