Finding \(\rm \dfrac{dy}{dx}\) for \(\rm x^m y^n = a^{m+n}\) using Implicit Differentiation

We are given the equation \(\rm x^m y^n = a^{m+n}\). We need to find \(\rm \dfrac{dy}{dx}\).

This equation involves both \(x\) and \(y\), and \(y\) is implicitly defined as a function of \(x\). We can use implicit differentiation to find \(\rm \dfrac{dy}{dx}\).

A helpful approach when dealing with products and powers is to take the natural logarithm of both sides of the equation before differentiating.

Given equation:

\(\rm x^m y^n = a^{m+n}\)

Take natural logarithm on both sides:

\(\ln(\rm x^m y^n) = \ln(a^{m+n})\)

Using the logarithm properties \(\ln(AB) = \ln A + \ln B\) and \(\ln(A^k) = k \ln A\):

\(\ln(\rm x^m) + \ln(y^n) = (m+n) \ln a\)

\(\rm m \ln x + n \ln y = (m+n) \ln a\)

Now, differentiate both sides with respect to \(x\). Remember that \(y\) is a function of \(x\), so we need to use the chain rule when differentiating terms involving \(y\).

\(\dfrac{d}{dx}(\rm m \ln x + n \ln y) = \dfrac{d}{dx}((m+n) \ln a)\)

Differentiate each term:

• For \(\rm m \ln x\): \(\dfrac{d}{dx}(\rm m \ln x) = m \cdot \dfrac{d}{dx}(\ln x) = m \cdot \dfrac{1}{x} = \dfrac{m}{x}\)
• For \(\rm n \ln y\): \(\dfrac{d}{dx}(\rm n \ln y) = n \cdot \dfrac{d}{dx}(\ln y)\). Since \(y\) is a function of \(x\), \(\dfrac{d}{dx}(\ln y) = \dfrac{1}{y} \cdot \dfrac{dy}{dx}\) (by chain rule). So, \(\dfrac{d}{dx}(\rm n \ln y) = n \cdot \dfrac{1}{y} \dfrac{dy}{dx} = \dfrac{n}{y} \dfrac{dy}{dx}\).
• For \(\rm (m+n) \ln a\): Since \(a\), \(m\), and \(n\) are constants, \(\rm (m+n) \ln a\) is also a constant. The derivative of a constant is 0. \(\dfrac{d}{dx}((m+n) \ln a) = 0\).

Substituting these derivatives back into the equation:

\(\dfrac{m}{x} + \dfrac{n}{y} \dfrac{dy}{dx} = 0\)

Now, we need to solve for \(\dfrac{dy}{dx}\). Isolate the term containing \(\dfrac{dy}{dx}\):

\(\dfrac{n}{y} \dfrac{dy}{dx} = -\dfrac{m}{x}\)

Multiply both sides by \(\dfrac{y}{n}\) to find \(\dfrac{dy}{dx}\):

\(\dfrac{dy}{dx} = -\dfrac{m}{x} \cdot \dfrac{y}{n}\)

\(\dfrac{dy}{dx} = -\dfrac{my}{nx}\)

Thus, the derivative \(\rm \dfrac{dy}{dx}\) is equal to \(-\rm \dfrac{my}{nx}\).