Find $$ \frac{d z}{d x} $$ and $$ \frac{d z}{d y} $$ for the following (A) $$ z=(4 x-2 y)(3 x+5 y) $$ (2) $$ z=\frac{5 x+y}{x-2 y} $$

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**(A) $$ z = (4x-2y)(3x+5y) $$**

1. Expand the product:
   $$
   z = (4x-2y)(3x+5y) = 4x\cdot3x + 4x\cdot5y -2y\cdot3x -2y\cdot5y = 12x^2 + 20xy - 6xy - 10y^2.
   $$
   Simplify:
   $$
   z = 12x^2 + 14xy - 10y^2.
   $$

2. Compute the partial derivative with respect to $$ x $$:
   $$
   \frac{dz}{dx} = \frac{\partial z}{\partial x} = \frac{d}{dx}(12x^2) + \frac{d}{dx}(14xy) + \frac{d}{dx}(-10y^2).
   $$
   Here, $$ y $$ is treated as a constant:
   $$
   \frac{dz}{dx} = 24x + 14y + 0 = 24x + 14y.
   $$

3. Compute the partial derivative with respect to $$ y $$:
   $$
   \frac{dz}{dy} = \frac{\partial z}{\partial y} = \frac{d}{dy}(12x^2) + \frac{d}{dy}(14xy) + \frac{d}{dy}(-10y^2).
   $$
   Here, $$ x $$ is treated as a constant:
   $$
   \frac{dz}{dy} = 0 + 14x - 20y = 14x - 20y.
   $$

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**(2) $$ z = \frac{5x + y}{x - 2y} $$**

Let $$ N = 5x + y $$ and $$ D = x - 2y $$.

1. Compute the partial derivative with respect to $$ x $$ using the quotient rule:
   $$
   \frac{dz}{dx} = \frac{N'_x \cdot D - N \cdot D'_x}{D^2},
   $$
   where:
   $$
   N'_x = \frac{d}{dx}(5x + y) = 5 \quad \text{(since $$ y $$ is constant)}
   $$
   and
   $$
   D'_x = \frac{d}{dx}(x - 2y) = 1.
   $$
   Substitute:
   $$
   \frac{dz}{dx} = \frac{5(x - 2y) - (5x + y)(1)}{(x-2y)^2} = \frac{5x - 10y - 5x - y}{(x-2y)^2} = \frac{-11y}{(x-2y)^2}.
   $$

2. Compute the partial derivative with respect to $$ y $$ using the quotient rule:
   $$
   \frac{dz}{dy} = \frac{N'_y \cdot D - N \cdot D'_y}{D^2},
   $$
   where:
   $$
   N'_y = \frac{d}{dy}(5x + y) = 1 \quad \text{(since $$ x $$ is constant)}
   $$
   and
   $$
   D'_y = \frac{d}{dy}(x - 2y) = -2.
   $$
   Substitute:
   $$
   \frac{dz}{dy} = \frac{1\cdot (x-2y) - (5x + y)(-2)}{(x-2y)^2} = \frac{x-2y + 10x + 2y}{(x-2y)^2} = \frac{11x}{(x-2y)^2}.
   $$

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**Final Answers:**

1. For $$ z = (4x-2y)(3x+5y) $$:
   $$
   \frac{dz}{dx} = 24x + 14y \quad \text{and} \quad \frac{dz}{dy} = 14x - 20y.
   $$

2. For $$ z = \frac{5x + y}{x - 2y} $$:
   $$
   \frac{dz}{dx} = \frac{-11y}{(x-2y)^2} \quad \text{and} \quad \frac{dz}{dy} = \frac{11x}{(x-2y)^2}.
   $$
For $$ z = (4x-2y)(3x+5y) $$: - $$ \frac{dz}{dx} = 24x + 14y $$ - $$ \frac{dz}{dy} = 14x - 20y $$ For $$ z = \frac{5x + y}{x - 2y} $$: - $$ \frac{dz}{dx} = \frac{-11y}{(x-2y)^2} $$ - $$ \frac{dz}{dy} = \frac{11x}{(x-2y)^2} $$

Find \( \frac{d z}{d x} \) and \( \frac{d z}{d y} \) for the following (A) \( z=(4 x-2 y)(3 x+5 y) \) (2) \( z=\frac{5 x+y}{x-2 y} \)

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