Partial Derivatives 6A6731

1. **State the problem:** Find the first partial derivatives $\frac{\partial h}{\partial x}$ and $\frac{\partial h}{\partial z}$ of the function $$h(x,z) = \sqrt{\frac{9xz}{3x - 4z}}.$$ 2. **Rewrite the function:** Let $$u = \frac{9xz}{3x - 4z},$$ so that $$h = \sqrt{u} = u^{\frac{1}{2}}.$$ 3. **Use the chain rule:** $$\frac{\partial h}{\partial x} = \frac{1}{2} u^{-\frac{1}{2}} \frac{\partial u}{\partial x}, \quad \frac{\partial h}{\partial z} = \frac{1}{2} u^{-\frac{1}{2}} \frac{\partial u}{\partial z}.$$ 4. **Find $\frac{\partial u}{\partial x}$:** $$u = \frac{9xz}{3x - 4z}.$$ Use the quotient rule: $$\frac{\partial u}{\partial x} = \frac{(9z)(3x - 4z) - (9xz)(3)}{(3x - 4z)^2}.$$ Simplify numerator: $$9z(3x - 4z) - 27xz = 27xz - 36z^2 - 27xz = -36z^2.$$ So, $$\frac{\partial u}{\partial x} = \frac{-36z^2}{(3x - 4z)^2}.$$ 5. **Find $\frac{\partial u}{\partial z}$:** Again, quotient rule: $$\frac{\partial u}{\partial z} = \frac{9x(3x - 4z) - 9xz(-4)}{(3x - 4z)^2}.$$ Simplify numerator: $$9x(3x - 4z) + 36xz = 27x^2 - 36xz + 36xz = 27x^2.$$ So, $$\frac{\partial u}{\partial z} = \frac{27x^2}{(3x - 4z)^2}.$$ 6. **Write the partial derivatives of $h$:** $$\frac{\partial h}{\partial x} = \frac{1}{2} u^{-\frac{1}{2}} \cdot \frac{-36z^2}{(3x - 4z)^2} = \frac{1}{2} \frac{-36z^2}{(3x - 4z)^2 \sqrt{u}} = \frac{-18z^2}{(3x - 4z)^2 \sqrt{\frac{9xz}{3x - 4z}}}.$$ $$\frac{\partial h}{\partial z} = \frac{1}{2} u^{-\frac{1}{2}} \cdot \frac{27x^2}{(3x - 4z)^2} = \frac{27x^2}{2(3x - 4z)^2 \sqrt{\frac{9xz}{3x - 4z}}}.$$ 7. **Final answers:** $$\boxed{\frac{\partial h}{\partial x} = \frac{-18z^2}{(3x - 4z)^2 \sqrt{\frac{9xz}{3x - 4z}}}}, \quad \boxed{\frac{\partial h}{\partial z} = \frac{27x^2}{2(3x - 4z)^2 \sqrt{\frac{9xz}{3x - 4z}}}}.$$