The aim of this question is to find the **first-order partial derivatives** of an **implicit** function made up of two **independent variables.**

The basis for this solution resolves around the **quotient rule of derivatives**. It states that if **$u$** and **$v$** are two functions, then the derivative of the **quotient $\frac{u}{v}$** can be calculated using the following formula:

\[\frac{d}{dx} \bigg ( \frac{u}{v} \bigg ) = \frac{v \cdot \frac{d}{dx}(u) – u \cdot \frac{d}{dx}(v)}{v^2}\]

Since there are **two independent** variables, there are **two parts to this question**. The first part calculates the **partial derivative** of **$f(x,y)$** with respect to variable **$x$** while second part calculates the **partial derivative** of **$f(x,y)$** with respect to variable **$y$**.

## Expert Answer

**Part 1: Calculating the partial derivative $\frac{\partial f(x,y)}{\partial x}$.**

\[ \frac{\partial f(x,y)}{\partial x} = \frac{\partial}{\partial x} \bigg (\frac{ax + by}{cx + dy}\bigg)\]

Applying the **quotient rule of derivatives**, we get:

\[ \frac{\partial f(x,y)}{\partial x} = \frac{(cx + dy)\frac{\partial}{\partial x}(ax + by) – (ax + by)\frac{\partial}{\partial x}(cx + dy)}{(cx + dy)^2}\]

Since we are calculating the **partial derivative** of **$f(x,y)$** with respect to **$x$**, the other independent variable **$y$ is being treated as a constant**.

Hence, **$\frac{\partial}{\partial x}(ax + by) = a$** and **$\frac{\partial}{\partial x}(cx + dy) = c$**. So the above expression reduces to following:

\[ \frac{\partial f(x,y)}{\partial x} = \frac{(cx + dy)(a)-(ax + by)(c)}{(cx + dy)^2}\]

\[ \frac{\partial f(x,y)}{\partial x} = \frac{acx + ady-(acx + bcy)}{(cx + dy)^2}\]

\[ \frac{\partial f(x,y)}{\partial x} = \frac{acx + ady – acx – bcy}{(cx + dy)^2}\]

\[ \frac{\partial f(x,y)}{\partial x} = \frac{ady – bcy}{(cx + dy)^2}\]

**\[ \frac{\partial f(x,y)}{\partial x} = \frac{(ad – bc)y}{(cx + dy)^2}\]**

**Part 2: Calculating the partial derivative $\frac{\partial f(x,y)}{\partial y}$.**

\[ \frac{\partial f(x,y)}{\partial y} = \frac{\partial}{\partial y} \bigg (\frac{ax + by}{cx + dy}\bigg)\]

Applying the **quotient rule of derivatives**, we get:

\[ \frac{\partial f(x,y)}{\partial y} = \frac{(cx + dy)\frac{\partial}{\partial y}(ax + by)-(ax + by)\frac{\partial}{\partial y}(cx + dy)}{(cx + dy)^2}\]

Since we are calculating the **partial derivative** of **$f(x,y)$** with respect to **$y$**, the other **independent** variable **$x$ is being treated as a constant**.

Hence, **$\frac{\partial}{\partial y}(ax + by) = b$** and **$\frac{\partial}{\partial y}(cx + dy) = d$**. So the above expression reduces to following:

\[ \frac{\partial f(x,y)}{\partial y} = \frac{(cx + dy)(b)-(ax + by)(d)}{(cx + dy)^2}\]

\[ \frac{\partial f(x,y)}{\partial y} = \frac{bcx + bdy-(adx + bdy)}{(cx + dy)^2}\]

\[ \frac{\partial f(x,y)}{\partial y} = \frac{bcx + bdy – adx – bdy}{(cx + dy)^2}\]

\[ \frac{\partial f(x,y)}{\partial y} = \frac{bcx – adx}{(cx + dy)^2}\]

## Numerical Result

The first **partial derivative** of the function is:

**\[ \frac{\partial f(x,y)}{\partial y} = \frac{(bc – ad)x}{(cx + dy)^2}\]**

## Example

Find the first **partial derivative** of the function $f(x,y) = \frac{2x + 4y}{6x + 8y}$ with respect to $x$.

\[ \frac{\partial f(x,y)}{\partial x} = \frac{(ad – bc)y}{(cx + dy)}^2 \]

\[ \frac{\partial f(x,y)}{\partial x} = \frac{[(2)(8) – (4)(6)]y}{(6)x + (8)y)^2} \]

\[ \frac{\partial f(x,y)}{\partial x} = -\frac{8y}{(6x + 8y)^2} \]