The given function is \( y = \sin(\cos(\tan(\cot x))) \) and we need to find the derivative of this function with respect to \( x \).
Solution:
To find the derivative \(\displaystyle \frac{dy}{dx}\) of the function \( y = \sin(\cos(\tan(\cot x))) \), we’ll use the chain rule. Here’s the step-by-step solution:
Let’s define the nested functions:
\( \begin{aligned} u &= \cot x \\ v &= \tan u = \tan(\cot x) \\ w &= \cos v = \cos(\tan(\cot x)) \\ y &= \sin w = \sin(\cos(\tan(\cot x))) \\ \end{aligned} \)Therefore, using the chain rule:
\( \displaystyle \frac{dy}{dx} = \frac{dy}{dw} \cdot \frac{dw}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx} \)Now, compute each derivative individually:
\( \begin{aligned} &\frac{dy}{dw} = \frac{d (\sin w)}{dw} = \cos w = \cos(\cos(\tan(\cot x))) \\ &\frac{dw}{dv} = \frac{d (\cos v)}{dv} = -\sin v = -\sin(\tan(\cot x)) \\ &\frac{dv}{du} = \frac{d (\tan u)}{du} = \sec^2 u = \sec^2 (\cot x) \\ &\frac{du}{dx} = \frac{d (\cot x)}{dx} = -\csc^2 x \quad (\csc \text{is} \operatorname{cosec})\\ \end{aligned} \)Multiply all the derivatives together:
\( \begin{aligned} \frac{dy}{dx} &= \cos(\cos(\tan(\cot x))) \cdot \left(-\sin(\tan(\cot x))\right) \cdot \sec^2 (\cot x) \cdot \left(-\csc^2 x\right) \\ &= \cos(\cos(\tan(\cot x))) \cdot \sin(\tan(\cot x)) \cdot \sec^2 (\cot x) \cdot \csc^2 x \\ \end{aligned} \)Thus, the derivative of the function is:
\( \displaystyle \boxed{\frac{dy}{dx} = \csc^2 x \cdot \sec^2 (\cot x) \cdot \sin(\tan(\cot x)) \cdot \cos(\cos(\tan(\cot x)))} \\ \)Please let me know in the comments if you find any errors in this solution.