Problem 19 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$s(y, z)=z^{2} \tan y z$$

Step-by-Step Solution

Answer

Answer: The first partial derivatives of the function are: $$\frac{\partial s}{\partial y} = z^3\sec^2(yz)$$ $$\frac{\partial s}{\partial z} = 2yz^2\sec^2(yz) + 2z\tan(yz)$$

Step 1: Find the partial derivative of s with respect to y

To find the partial derivative of s(y, z) with respect to y, we can treat z as a constant and differentiate the function with respect to y. $$\frac{\partial s}{\partial y} =\frac{d}{dy}\left(z^{2} \tan y z\right)$$ To do this differentiation, we will need to use the chain rule: $$\frac{d}{dy}(\tan(uz)) = (\frac{d}{du}\tan(u)) \cdot\frac{d(uz)}{dy}$$ Here, let \(u = yz\). So, the partial derivative will be: $$\frac{\partial s}{\partial y} = z^2\frac{d}{dy}(\tan(yz))$$ $$\frac{\partial s}{\partial y} = z^2\cdot (\frac{d}{d u}\tan(u)) \cdot \frac{d(yz)}{dy}$$ $$\frac{\partial s}{\partial y} = z^2 \cdot (\sec^2(u)) \cdot z = z^3\sec^2(yz)$$

Step 2: Find the partial derivative of s with respect to z

To find the partial derivative of s(y, z) with respect to z, we treat y as constant and differentiate the function with respect to z. $$ \frac{\partial s}{\partial z} = \frac{d}{dz}\left(z^{2} \tan y z\right) $$ First, we apply product rule on \(z^2 \tan(yz)\): $$\frac{d}{dz}(u v) = u \frac{d}{dz}v + v \frac{d}{dz}u$$ Here, let \(u=z^2\) and \(v=\tan(yz)\). So, we will have: $$\frac{d}{dz}u = 2z$$ $$\frac{d}{dz}v = \frac{d}{d(u)}(\tan(u)) \cdot \frac{d(yu)}{dz} = \sec^2(u) \cdot y$$ Now, we can apply the product rule formula: $$\frac{\partial s}{\partial z} = z^2 \cdot \sec^2(yz) \cdot y + \tan(yz) \cdot 2z$$ $$\frac{\partial s}{\partial z} = 2yz^2\sec^2(yz) + 2z\tan(yz)$$ So, the first partial derivatives are: $$\frac{\partial s}{\partial y} = z^3\sec^2(yz)$$ $$\frac{\partial s}{\partial z} = 2yz^2\sec^2(yz) + 2z\tan(yz)$$