# Y = 2x^3sinx - 3xcosx Find the derivative of the equation?

Mar 4, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 {x}^{3} - 3\right) \cos x + \left(6 {x}^{2} + 3 x\right) \sin x$

#### Explanation:

Each term has to be differentiated using the $\textcolor{b l u e}{\text{product rule}}$

$\text{Given "f(x)=g(x).h(x)" then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}} \leftarrow \text{ product rule}$

$\textcolor{m a \ge n t a}{\text{First term}}$

$\text{for } f \left(x\right) = 2 {x}^{3} \sin x$

$\text{here } g \left(x\right) = 2 {x}^{3} \Rightarrow g ' \left(x\right) = 6 {x}^{2}$

$h \left(x\right) = \sin x \Rightarrow h ' \left(x\right) = \cos x$

$\Rightarrow f ' \left(x\right) = 2 {x}^{3} \left(\cos x\right) + 6 {x}^{2} \left(\sin x\right) \to \left(1\right)$

$\textcolor{m a \ge n t a}{\text{Second term}}$

$f \left(x\right) = 3 x \cos x$

$\text{here } g \left(x\right) = 3 x \Rightarrow g ' \left(x\right) = 3$

$h \left(x\right) = \cos x \Rightarrow h ' \left(x\right) = - \sin x$

$\Rightarrow f ' \left(x\right) = 3 x \left(- \sin x\right) + 3 \cos x \to \left(2\right)$

$\text{Combining differentiated terms, that is } \left(1\right) - \left(2\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{3} \cos x + 6 {x}^{2} \sin x + 3 x \sin x - 3 \cos x$

.>$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 {x}^{3} - 3\right) \cos x + \left(6 {x}^{2} + 3 x\right) \sin x$