Differentiating the Sine Function
Find \(y^\prime\) if:
- \(y=x+4\sin x\)
- \(y=x^{2}\sin x\)
- \(y=\dfrac{\sin x}{x}\)
- \(y=e^{x}\sin x\)
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
- \(y^\prime =\dfrac{d}{dx}\left(\dfrac{\sin x}{x}\right)\) = \(\dfrac{\left[\dfrac{d}{dx}\sin x\right] \cdot x-\sin x\cdot \left[ \dfrac{d }{dx}x\right]}{x^{2}}=\dfrac{x\cos x-\sin x}{x^{2}}\)
- \[ \begin{eqnarray*} y^\prime &=& \dfrac{d}{dx}(e^{x}\sin x) = (e^{x} \dfrac{d}{dx}\sin x + \left(\dfrac{d}{dx}e^{x}\right) \sin x = e^{x}\cos x+e^{x}\sin x\\ &=& e^{x} (\cos x+\sin x) \end{eqnarray*} \]