Question

Find the derivative of $\left(3x^2+5\right)\sin x$.

Answer 1

Apply the formula to calculate the derivative

The derivative of the product of two functions (say) $u$ and $v$ is defined as follows,

$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(u,v\right)\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\left(\frac{d}{dx}v\right)\mathbf{}\mathbf{+}\mathbf{}\mathit{v}\left(\frac{d}{dx}u\right)$

Then, the derivative of $\left(3x^2+5\right)\sin x$ can be calculated as,

$\frac{\mathrm{d}}{\mathrm{dx}}\left[\left(3{\mathrm{x}}^2+5\right)\mathrm{sinx}\right]=\left(3{\mathrm{x}}^2+5\right)\left(\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\right)+\mathrm{sinx}\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(3{\mathrm{x}}^2+5\right)\right]$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\left(3{\mathrm{x}}^2+5\right)\mathrm{sinx}\right]=\left(3{\mathrm{x}}^2+5\right)\cos x+\mathrm{sinx}\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(3{\mathrm{x}}^2+5\right)\right]$ $\left[\because \frac{d}{dx}sinx=cosx\right]$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\left(3{\mathrm{x}}^2+5\right)\mathrm{sinx}\right]=\left(3{\mathrm{x}}^2+5\right)\cos x+\mathrm{sinx}\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(3{\mathrm{x}}^2\right)+\frac{d}{dx}5\right]$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\left(3{\mathrm{x}}^2+5\right)\mathrm{sinx}\right]=\left(3{\mathrm{x}}^2+5\right)\cos x+\mathrm{sinx}\left[3\left(\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^2\right)+\frac{d}{dx}5\right]$ $\left[\because \frac{d}{dx}\left(n{x}^n\right)=n\left(\frac{d}{dx}{x}^n\right),\mathrm{where}n=\mathrm{constant}\right]$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\left(3{\mathrm{x}}^2+5\right)\mathrm{sinx}\right]=\left(3{\mathrm{x}}^2+5\right)\cos x+\mathrm{sinx}\left[3\left(2\mathrm{x}\right)+0\right]$ $[\because \frac{d}{dx}{x}^n=n{x}^{n-1},\frac{d}{dx}C=0]$

$$\begin{gathered} \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\left(3{\mathrm{x}}^2+5\right)\mathrm{sinx}\right]=\left(3{\mathrm{x}}^2+5\right)\cos x+\mathrm{sinx}\left[6x\right] \\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\left(3{\mathrm{x}}^2+5\right)\mathrm{sinx}\right]=\left(3{\mathrm{x}}^2+5\right)\cos x+6x\sin x \end{gathered}$$

Hence, the derivative of $\left(3x^2+5\right)\sin x$ is $\left(3x^2+5\right)\cos x+6x\sin x$.