Figure

Find $y^\prime$ if:

Solution

$y^\prime =\dfrac{d}{dx}(x+4\sin x)$ = $\dfrac{d}{dx}x+\dfrac{d}{dx}(4 \sin x)$ = 1+4 $\dfrac{d}{dx} \sin x$ = 1 + 4 $\cos x$
$y^\prime = \dfrac{d}{dx}(x^{2} \sin x)$ = $x^{2} \left[\dfrac{d}{dx} \sin x\right] +\left[\dfrac{d}{dx}x^{2}\right] \sin x$ = $x^{2}\cos x+2x\sin x$
187
$y^\prime =\dfrac{d}{dx}\left(\dfrac{\sin x}{x}\right)$ = $\dfrac{\left[\dfrac{d}{dx}\sin x\right] \cdot x-\sin x\cdot \left[ \dfrac{d }{dx}x\right]}{x^{2}}=\dfrac{x\cos x-\sin x}{x^{2}}$
$\begin{eqnarray*} y^\prime &=& \dfrac{d}{dx}(e^{x}\sin x) = (e^{x} \dfrac{d}{dx}\sin x + \left(\dfrac{d}{dx}e^{x}\right) \sin x = e^{x}\cos x+e^{x}\sin x\\ &=& e^{x} (\cos x+\sin x) \end{eqnarray*}$