Problem 18 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$g(x, z)=x \ln \left(z^{2}+x^{2}\right)$$

Step-by-Step Solution

Answer

Question: Compute the first partial derivatives of the function \(g(x, z)=x \ln \left(z^{2}+x^{2}\right)\) with respect to the variables \(x\) and \(z\). Answer: The first partial derivatives of the function \(g(x, z)\) are: \(\frac{\partial g}{\partial x} = \ln \left(z^{2}+x^{2}\right) + \frac{2x^{2}}{z^{2}+x^{2}}\) \(\frac{\partial g}{\partial z} = \frac{2xz}{z^{2}+x^{2}}\)

Step 1: Compute the first partial derivative with respect to \(x\)

We need to find \(\frac{\partial g}{\partial x}\). To do this, we can apply the product rule for differentiation. The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. So, to differentiate \(g(x, z)=x \ln \left(z^{2}+x^{2}\right)\) with respect to \(x\), we have: $$\frac{\partial g}{\partial x} = \frac{d(x)}{dx} \cdot \ln \left(z^{2}+x^{2}\right) +x \cdot \frac{d}{dx}\left[\ln \left(z^{2}+x^{2}\right)\right]$$ Now, find the derivative of each part: $$\frac{d(x)}{dx} = 1$$ $$\frac{d}{dx}\left[\ln \left(z^{2}+x^{2}\right)\right] = \frac{1}{z^{2}+x^{2}}\cdot\frac{d}{dx}\left[z^{2}+x^{2}\right] = \frac{2x}{z^{2}+x^{2}}$$ Finally, substitute the derivatives back into the equation: $$\frac{\partial g}{\partial x} = 1 \cdot \ln \left(z^{2}+x^{2}\right) + x \cdot \frac{2x}{z^{2}+x^{2}}$$ Simplify the equation: $$\frac{\partial g}{\partial x} = \ln \left(z^{2}+x^{2}\right) + \frac{2x^{2}}{z^{2}+x^{2}}$$

Step 2: Compute the first partial derivative with respect to \(z\)

We need to find \(\frac{\partial g}{\partial z}\). To do this, we can apply the chain rule for differentiation. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. So, to differentiate \(g(x, z)=x \ln \left(z^{2}+x^{2}\right)\) with respect to \(z\), we have: $$\frac{\partial g}{\partial z} = x \cdot \frac{d}{dz}\left[\ln \left(z^{2}+x^{2}\right)\right]$$ Now, find the derivative of the function inside the logarithm with respect to \(z\): $$\frac{d}{dz}\left[z^{2}+x^{2}\right] = 2z$$ Substitute the derivative back into the equation: $$\frac{\partial g}{\partial z} = x \cdot \frac{1}{z^{2}+x^{2}}\cdot 2z$$ Simplify the equation: $$\frac{\partial g}{\partial z} = \frac{2xz}{z^{2}+x^{2}}$$ The final results for the first partial derivatives are: $$\frac{\partial g}{\partial x} = \ln \left(z^{2}+x^{2}\right) + \frac{2x^{2}}{z^{2}+x^{2}}$$ $$\frac{\partial g}{\partial z} = \frac{2xz}{z^{2}+x^{2}}$$