Problem 20 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$F(p, q)=\sqrt{p^{2}+p q+q^{2}}$$

Step-by-Step Solution

Answer

Question: Find the first partial derivatives of the function \(F(p, q) = \sqrt{p^{2} + pq + q^{2}}\). Answer: The first partial derivatives of the function are: 1. With respect to \(p\): \(\frac{\partial F}{\partial p} = \frac{1}{2\sqrt{p^{2}+p q+q^{2}}} (2p + q)\) 2. With respect to \(q\): \(\frac{\partial F}{\partial q} = \frac{1}{2\sqrt{p^{2}+p q+q^{2}}} (p + 2q)\)

Step 1: Write the given function

The given function is: $$F(p, q)=\sqrt{p^{2}+p q+q^{2}}$$

Step 2: Find the partial derivative with respect to \(p\)

To find the partial derivative of \(F(p,q)\) with respect to \(p\), we'll treat \(q\) as a constant: $$\frac{\partial F}{\partial p} = \frac{\partial}{\partial p} \sqrt{p^{2}+p q+q^{2}}$$ We use the chain rule, where \(\frac{dF}{dp} = \frac{dF}{dz}\frac{dz}{dp}\), and let \(z = p^{2}+p q+q^{2}\). Then: $$\frac{\partial F}{\partial p} = \frac{1}{2\sqrt{z}}\frac{\partial z}{\partial p}$$ Now we differentiate \(z\) with respect to \(p\): $$\frac{\partial z}{\partial p} = \frac{\partial}{\partial p} \left( p^{2} + p q + q^{2} \right) = 2p + q$$ Finally, plug this back into our expression for \(\frac{\partial F}{\partial p}\): $$\frac{\partial F}{\partial p} = \frac{1}{2\sqrt{p^{2}+p q+q^{2}}} (2p + q)$$

Step 3: Find the partial derivative with respect to \(q\)

Now, we need to find the partial derivative of \(F(p,q)\) with respect to \(q\), this time treating \(p\) as a constant: $$\frac{\partial F}{\partial q} = \frac{\partial}{\partial q} \sqrt{p^{2}+p q+q^{2}}$$ Again, we use the chain rule with \(z = p^{2}+p q+q^{2}\): $$\frac{\partial F}{\partial q} = \frac{1}{2\sqrt{z}}\frac{\partial z}{\partial q}$$ Now we differentiate \(z\) with respect to \(q\): $$\frac{\partial z}{\partial q} = \frac{\partial}{\partial q} \left( p^{2} + p q + q^{2} \right) = p + 2q$$ Finally, plug this back into our expression for \(\frac{\partial F}{\partial q}\): $$\frac{\partial F}{\partial q} = \frac{1}{2\sqrt{p^{2}+p q+q^{2}}} (p + 2q)$$

Step 4: Write down the first partial derivatives

We've found the first partial derivatives of \(F(p,q)\). They are: 1. With respect to \(p\): \(\frac{\partial F}{\partial p} = \frac{1}{2\sqrt{p^{2}+p q+q^{2}}} (2p + q)\) 2. With respect to \(q\): \(\frac{\partial F}{\partial q} = \frac{1}{2\sqrt{p^{2}+p q+q^{2}}} (p + 2q)\)