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Find the first partial derivatives of the function f(x, y) = (ax + by)/(cx + dy)

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 $x$.

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

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

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

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