The derivative of $\sin ( \theta \cos \theta )$ is $\cos ( \theta \cos \theta ) \cdot [ \cos \theta - \theta \sin \theta ]$ .

Question

Quizplus · Accepted Answer

This is correct because by applying the chain rule for differentiation, the derivative of $\sin(u)$ with respect to $\theta$, where $u = \theta \cos \theta$, is $\cos(u) \cdot \frac{du}{d\theta}$. Here, $\frac{du}{d\theta} = \cos \theta - \theta \sin \theta$, leading to the given expression.

The Derivative Of sin⁡(θcos⁡θ)\sin ( \theta \cos \theta )sin(θcosθ) Is cos⁡(θcos⁡θ)⋅[cos⁡θ−θsin⁡θ]\cos ( \theta \cos \theta ) \cdot [ \cos \theta - \theta \sin \theta ]cos(θcosθ)⋅[cosθ−θsinθ]

The Derivative Of $\sin ( \theta \cos \theta )$ Is $\cos ( \theta \cos \theta ) \cdot [ \cos \theta - \theta \sin \theta ]$