1. The problem asks to evaluate the expression $\left(\frac{\partial f}{\partial v} + \frac{\partial f}{\partial x} = \frac{\partial f}{\partial w} + 7 \frac{\partial f}{\partial v}\right) \left(\frac{45 \frac{\partial f}{\partial x}}{3 \frac{\partial f}{\partial x}} - \frac{\partial f}{\partial z}\right)$ for the given function $$f(u,v,w,x,y,z) = \frac{7u^2 x^5 z + 3z^2 u w - \frac{7y^5}{x}}{\frac{7x^2}{4} w + 7 w^5} - \left(\frac{4 w^3 x^3}{w x} - 7 a s^2\right) - \sqrt{7} v y z.$$ 2. We need to find the partial derivatives $\frac{\partial f}{\partial v}$, $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial w}$, and $\frac{\partial f}{\partial z}$. 3. First, note that $f$ is composed of three parts: - Part 1: $$\frac{7u^2 x^5 z + 3z^2 u w - \frac{7y^5}{x}}{\frac{7x^2}{4} w + 7 w^5}$$ - Part 2: $$\frac{4 w^3 x^3}{w x} - 7 a s^2$$ - Part 3: $$- \sqrt{7} v y z$$ 4. Calculate $\frac{\partial f}{\partial v}$: - Part 1 and Part 2 do not depend on $v$, so their derivatives are zero. - Part 3 derivative w.r.t. $v$ is $$\frac{\partial}{\partial v}(- \sqrt{7} v y z) = - \sqrt{7} y z.$$ So, $$\frac{\partial f}{\partial v} = - \sqrt{7} y z.$$ 5. Calculate $\frac{\partial f}{\partial x}$: - For Part 1, use quotient rule: if $f = \frac{N}{D}$, then $$\frac{\partial f}{\partial x} = \frac{D \frac{\partial N}{\partial x} - N \frac{\partial D}{\partial x}}{D^2}.$$ - Let $$N = 7u^2 x^5 z + 3z^2 u w - \frac{7 y^5}{x}, \quad D = \frac{7 x^2}{4} w + 7 w^5.$$ - Compute $\frac{\partial N}{\partial x}$: $$\frac{\partial}{\partial x}(7u^2 x^5 z) = 7 u^2 z \cdot 5 x^4 = 35 u^2 z x^4,$$ $$\frac{\partial}{\partial x}(3 z^2 u w) = 0,$$ $$\frac{\partial}{\partial x}\left(- \frac{7 y^5}{x}\right) = -7 y^5 \cdot \frac{\partial}{\partial x} \left(\frac{1}{x}\right) = -7 y^5 (-\frac{1}{x^2}) = \frac{7 y^5}{x^2}.$$ So, $$\frac{\partial N}{\partial x} = 35 u^2 z x^4 + \frac{7 y^5}{x^2}.$$ - Compute $\frac{\partial D}{\partial x}$: $$\frac{\partial}{\partial x} \left(\frac{7 x^2}{4} w + 7 w^5\right) = \frac{7}{4} w \cdot 2 x = \frac{7}{2} w x.$$ - Therefore, $$\frac{\partial f}{\partial x} = \frac{\left(\frac{7 x^2}{4} w + 7 w^5\right) \left(35 u^2 z x^4 + \frac{7 y^5}{x^2}\right) - \left(7 u^2 x^5 z + 3 z^2 u w - \frac{7 y^5}{x}\right) \left(\frac{7}{2} w x\right)}{\left(\frac{7 x^2}{4} w + 7 w^5\right)^2} - \frac{\partial}{\partial x} \left(\frac{4 w^3 x^3}{w x} - 7 a s^2\right) - 0,$$ since Part 3 does not depend on $x$. - Simplify Part 2 derivative: $$\frac{4 w^3 x^3}{w x} = 4 w^3 x^3 \cdot \frac{1}{w x} = 4 w^{3-1} x^{3-1} = 4 w^2 x^2,$$ so $$\frac{\partial}{\partial x} (4 w^2 x^2) = 4 w^2 \cdot 2 x = 8 w^2 x,$$ and derivative of $-7 a s^2$ w.r.t. $x$ is zero. - Thus, $$\frac{\partial f}{\partial x} = \frac{\left(\frac{7 x^2}{4} w + 7 w^5\right) \left(35 u^2 z x^4 + \frac{7 y^5}{x^2}\right) - \left(7 u^2 x^5 z + 3 z^2 u w - \frac{7 y^5}{x}\right) \left(\frac{7}{2} w x\right)}{\left(\frac{7 x^2}{4} w + 7 w^5\right)^2} - 8 w^2 x.$$ 6. Calculate $\frac{\partial f}{\partial w}$: - For Part 1, again use quotient rule with respect to $w$. - Compute $\frac{\partial N}{\partial w}$: $$\frac{\partial}{\partial w} (7 u^2 x^5 z) = 0,$$ $$\frac{\partial}{\partial w} (3 z^2 u w) = 3 z^2 u,$$ $$\frac{\partial}{\partial w} \left(- \frac{7 y^5}{x}\right) = 0,$$ so $$\frac{\partial N}{\partial w} = 3 z^2 u.$$ - Compute $\frac{\partial D}{\partial w}$: $$\frac{\partial}{\partial w} \left(\frac{7 x^2}{4} w + 7 w^5\right) = \frac{7 x^2}{4} + 35 w^4.$$ - Therefore, $$\frac{\partial f}{\partial w} = \frac{\left(\frac{7 x^2}{4} w + 7 w^5\right) (3 z^2 u) - \left(7 u^2 x^5 z + 3 z^2 u w - \frac{7 y^5}{x}\right) \left(\frac{7 x^2}{4} + 35 w^4\right)}{\left(\frac{7 x^2}{4} w + 7 w^5\right)^2} - \frac{\partial}{\partial w} \left(4 w^2 x^2 - 7 a s^2\right) - 0,$$ since Part 3 does not depend on $w$. - Derivative of Part 2 w.r.t. $w$: $$\frac{\partial}{\partial w} (4 w^2 x^2) = 8 w x^2,$$ and derivative of $-7 a s^2$ w.r.t. $w$ is zero. - So, $$\frac{\partial f}{\partial w} = \frac{\left(\frac{7 x^2}{4} w + 7 w^5\right) (3 z^2 u) - \left(7 u^2 x^5 z + 3 z^2 u w - \frac{7 y^5}{x}\right) \left(\frac{7 x^2}{4} + 35 w^4\right)}{\left(\frac{7 x^2}{4} w + 7 w^5\right)^2} - 8 w x^2.$$ 7. Calculate $\frac{\partial f}{\partial z}$: - For Part 1, treat $z$ as variable: $$\frac{\partial N}{\partial z} = 7 u^2 x^5 + 6 z u w - 0 = 7 u^2 x^5 + 6 z u w,$$ since $-\frac{7 y^5}{x}$ does not depend on $z$. - $D$ does not depend on $z$, so $$\frac{\partial f}{\partial z} = \frac{\frac{\partial N}{\partial z} D - N \cdot 0}{D^2} - 0 - \frac{\partial}{\partial z} (\sqrt{7} v y z) = \frac{(7 u^2 x^5 + 6 z u w) D}{D^2} - \sqrt{7} v y = \frac{7 u^2 x^5 + 6 z u w}{D} - \sqrt{7} v y.$$ 8. Now substitute these into the expression: $$\left(\frac{\partial f}{\partial v} + \frac{\partial f}{\partial x} = \frac{\partial f}{\partial w} + 7 \frac{\partial f}{\partial v}\right) \left(\frac{45 \frac{\partial f}{\partial x}}{3 \frac{\partial f}{\partial x}} - \frac{\partial f}{\partial z}\right).$$ - Simplify the first parenthesis: $$\frac{\partial f}{\partial v} + \frac{\partial f}{\partial x} = \frac{\partial f}{\partial w} + 7 \frac{\partial f}{\partial v} \implies \frac{\partial f}{\partial x} - \frac{\partial f}{\partial w} = 6 \frac{\partial f}{\partial v}.$$ - So the first parenthesis equals zero if the equality holds, or rearranged as above. - The second parenthesis: $$\frac{45 \frac{\partial f}{\partial x}}{3 \frac{\partial f}{\partial x}} - \frac{\partial f}{\partial z} = 15 - \frac{\partial f}{\partial z}.$$ 9. Without specific values for variables, the expression cannot be numerically evaluated further. Hence, the evaluated expression simplifies to: $$\left(\frac{\partial f}{\partial x} - \frac{\partial f}{\partial w} - 6 \frac{\partial f}{\partial v}\right) \left(15 - \frac{\partial f}{\partial z}\right).$$