Problem 22 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$h(u, v)=\sqrt{\frac{u v}{u-v}}$$

Step-by-Step Solution

Answer

Question: Determine the first partial derivatives of the function \(h(u, v) = \sqrt{\frac{u v}{u - v}}\) with respect to both variables \(u\) and \(v\). Answer: The first partial derivatives are as follows: $$\frac{\partial h}{\partial u} = \frac{-v^2}{2(u-v)^2\sqrt{\frac{u v}{u-v}}},$$ $$\frac{\partial h}{\partial v} = \frac{u^2}{2(u-v)^2\sqrt{\frac{u v}{u-v}}}.$$

Step 1: Find the partial derivative with respect to u

To find the partial derivative of \(h(u, v)\) with respect to \(u\), we will consider \(v\) as a constant and apply the chain rule. $$\frac{\partial h}{\partial u}=\frac{\partial}{\partial u} \sqrt{\frac{u v}{u-v}}$$ Let $$f(u, v) = \frac{u v}{u - v}.$$ Then, \(h(u, v) = \sqrt{f(u, v)}\). Using chain rule, we can write, $$\frac{\partial h}{\partial u} = \frac{\mathrm{d}\sqrt{f(u, v)}}{\mathrm{d}f(u, v)}\frac{\partial f(u,v)}{\partial u}$$ Now, we first find the derivative of \(h(u, v)\) with respect to \(f(u, v)\): $$\frac{\mathrm{d}\sqrt{f(u, v)}}{\mathrm{d}f(u, v)} = \frac{1}{2\sqrt{f(u, v)}}$$ Next, we find the partial derivative of \(f(u, v)\) with respect to \(u\): $$\frac{\partial f(u,v)}{\partial u} = \frac{\partial}{\partial u}\left(\frac{u v}{u-v}\right)$$ Using quotient rule, $$\frac{\partial f(u,v)}{\partial u} = \frac{v(u-v) - u v}{(u-v)^2} = \frac{-v^2}{(u-v)^2}$$ Now, substituting the derived expressions, we have: $$\frac{\partial h}{\partial u} = \frac{1}{2\sqrt{f(u, v)}}\times\frac{-v^2}{(u-v)^2} = \frac{-v^2}{2(u-v)^2\sqrt{\frac{u v}{u-v}}}$$

Step 2: Find the partial derivative with respect to v

To find the partial derivative of \(h(u,v)\) with respect to \(v\), we will consider \(u\) as a constant and again apply the chain rule. $$\frac{\partial h}{\partial v}=\frac{\partial}{\partial v} \sqrt{\frac{u v}{u-v}}$$ Using chain rule, we can write, $$\frac{\partial h}{\partial v} = \frac{\mathrm{d}\sqrt{f(u, v)}}{\mathrm{d}f(u, v)}\frac{\partial f(u,v)}{\partial v}$$ We've already found the derivative of \(h(u, v)\) with respect to \(f(u, v)\) in Step 1: $$\frac{\mathrm{d}\sqrt{f(u, v)}}{\mathrm{d}f(u, v)} = \frac{1}{2\sqrt{f(u, v)}}$$ Now, we find the partial derivative of \(f(u, v)\) with respect to \(v\): $$\frac{\partial f(u,v)}{\partial v} = \frac{\partial}{\partial v}\left(\frac{u v}{u-v}\right)$$ Using quotient rule, $$\frac{\partial f(u,v)}{\partial v} = \frac{u^2 - u(u - v)}{(u-v)^2} = \frac{u^2}{(u-v)^2}$$ Now, substituting the derived expressions, we have: $$\frac{\partial h}{\partial v} = \frac{1}{2\sqrt{f(u, v)}}\times\frac{u^2}{(u-v)^2} = \frac{u^2}{2(u-v)^2\sqrt{\frac{u v}{u-v}}}$$ In conclusion, the first partial derivatives of the function \(h(u, v)\) with respect to \(u\) and \(v\) are: $$\frac{\partial h}{\partial u} = \frac{-v^2}{2(u-v)^2\sqrt{\frac{u v}{u-v}}}$$ $$\frac{\partial h}{\partial v} = \frac{u^2}{2(u-v)^2\sqrt{\frac{u v}{u-v}}}$$