# How do you find the derivative of sqrtθ sin θ?

Let $f \left(x\right) = \sqrt{\theta} \sin \theta$$= {\left(\theta\right)}^{\frac{1}{2}} \sin \theta$
differentiate with respect to $\theta$ using product rule
$\frac{\mathrm{df} \left(x\right)}{d \theta} = \frac{d {\left(\theta\right)}^{\frac{1}{2}}}{d \theta} \sin \theta + {\theta}^{\frac{1}{2}} \frac{d \left(\sin \theta\right)}{d \theta}$
$\implies \frac{\mathrm{df} \left(x\right)}{d \theta} = \frac{1}{2} {\theta}^{\frac{1}{2} - 1} \sin \theta + {\theta}^{\frac{1}{2}} \cos \theta$
$\implies \frac{\mathrm{df} \left(x\right)}{d \theta} = \frac{1}{2} {\theta}^{- \frac{1}{2}} \sin \theta + \sqrt{\theta} \cos \theta$
$\implies \frac{\mathrm{df} \left(x\right)}{d \theta} = \frac{1}{2} \left(\frac{1}{\theta} ^ \left(\frac{1}{2}\right)\right) \sin \theta + \sqrt{\theta} \cos \theta$
$\implies \frac{\mathrm{df} \left(x\right)}{d \theta} = \frac{1}{2} \left(\frac{1}{\sqrt{\theta}}\right) \sin \theta + \sqrt{\theta} \cos \theta$
$\implies \frac{\mathrm{df} \left(x\right)}{d \theta} = \frac{1}{2 \sqrt{\theta}} \sin \theta + \sqrt{\theta} \cos \theta$