1. **Problem Statement:** We have a vector function $$\mathbf{w} = \begin{pmatrix} 4t \\ r \\ s \end{pmatrix}$$ where $$t = \tan^{-1}\left(\frac{u}{v}\right), s = 2uv, r = \ln\left(\frac{v}{2u}\right).$$ We need to: - Determine the number of chains and terms in each chain. - Compute the chains in terms of variables. - Find the partial derivatives $$\frac{\partial \mathbf{w}}{\partial u}$$ and $$\frac{\partial \mathbf{w}}{\partial v}$$ at $$u=1$$ and $$v=1 + \text{last digit of registration number}$$. 2. **Number of Chains and Terms:** - The vector $$\mathbf{w}$$ has 3 components: $$4t$$, $$r$$, and $$s$$. - Each component depends on variables $$u$$ and $$v$$ through intermediate functions $$t$$, $$r$$, and $$s$$. - So, there are 3 chains, one for each component. - Each chain has 2 terms (variables $$u$$ and $$v$$) because $$t$$, $$r$$, and $$s$$ depend on both $$u$$ and $$v$$. 3. **Compute the Chains:** - We use the multivariate chain rule: $$\frac{\partial w_i}{\partial u} = \frac{\partial w_i}{\partial t} \frac{\partial t}{\partial u} + \frac{\partial w_i}{\partial r} \frac{\partial r}{\partial u} + \frac{\partial w_i}{\partial s} \frac{\partial s}{\partial u}$$ But since $$w_1 = 4t$$ depends only on $$t$$, $$w_2 = r$$ depends only on $$r$$, and $$w_3 = s$$ depends only on $$s$$, the partial derivatives simplify: - $$\frac{\partial w_1}{\partial u} = 4 \frac{\partial t}{\partial u}$$ - $$\frac{\partial w_2}{\partial u} = \frac{\partial r}{\partial u}$$ - $$\frac{\partial w_3}{\partial u} = \frac{\partial s}{\partial u}$$ Similarly for $$v$$. 4. **Calculate Partial Derivatives of $$t$$, $$r$$, and $$s$$:** - $$t = \tan^{-1}\left(\frac{u}{v}\right)$$ Using the derivative of $$\tan^{-1}(x)$$: $$\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2}$$ So, $$\frac{\partial t}{\partial u} = \frac{1}{1 + \left(\frac{u}{v}\right)^2} \cdot \frac{1}{v} = \frac{v}{v^2 + u^2}$$ $$\frac{\partial t}{\partial v} = \frac{1}{1 + \left(\frac{u}{v}\right)^2} \cdot \left(-\frac{u}{v^2}\right) = -\frac{u}{v^2 + u^2}$$ - $$s = 2uv$$ $$\frac{\partial s}{\partial u} = 2v$$ $$\frac{\partial s}{\partial v} = 2u$$ - $$r = \ln\left(\frac{v}{2u}\right) = \ln(v) - \ln(2u)$$ $$\frac{\partial r}{\partial u} = -\frac{1}{u}$$ $$\frac{\partial r}{\partial v} = \frac{1}{v}$$ 5. **Calculate $$\frac{\partial \mathbf{w}}{\partial u}$$ and $$\frac{\partial \mathbf{w}}{\partial v}$$:** $$\frac{\partial \mathbf{w}}{\partial u} = \begin{pmatrix} 4 \frac{\partial t}{\partial u} \\ \frac{\partial r}{\partial u} \\ \frac{\partial s}{\partial u} \end{pmatrix} = \begin{pmatrix} 4 \cdot \frac{v}{v^2 + u^2} \\ -\frac{1}{u} \\ 2v \end{pmatrix}$$ $$\frac{\partial \mathbf{w}}{\partial v} = \begin{pmatrix} 4 \frac{\partial t}{\partial v} \\ \frac{\partial r}{\partial v} \\ \frac{\partial s}{\partial v} \end{pmatrix} = \begin{pmatrix} 4 \cdot \left(-\frac{u}{v^2 + u^2}\right) \\ \frac{1}{v} \\ 2u \end{pmatrix}$$ 6. **Evaluate at $$u=1$$ and $$v=1 + d$$ where $$d$$ is the last digit of registration number:** Let $$v = 1 + d$$. Substitute: $$\frac{\partial \mathbf{w}}{\partial u} = \begin{pmatrix} \frac{4(1 + d)}{(1 + d)^2 + 1} \\ -1 \\ 2(1 + d) \end{pmatrix}$$ $$\frac{\partial \mathbf{w}}{\partial v} = \begin{pmatrix} -\frac{4}{(1 + d)^2 + 1} \\ \frac{1}{1 + d} \\ 2 \end{pmatrix}$$ **Final answers:** $$\boxed{\frac{\partial \mathbf{w}}{\partial u} = \begin{pmatrix} \frac{4(1 + d)}{(1 + d)^2 + 1} \\ -1 \\ 2(1 + d) \end{pmatrix}, \quad \frac{\partial \mathbf{w}}{\partial v} = \begin{pmatrix} -\frac{4}{(1 + d)^2 + 1} \\ \frac{1}{1 + d} \\ 2 \end{pmatrix}}$$