If \(\rm x^m y^n =a^{m+n}\) , then what is \(\rm \dfrac{dy}{dx}\) equal to?
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 \(-\rm \dfrac{my}{nx}\).
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Important Questions from Evaluation of derivatives
What is the derivative of sin(ln x) + cos(ln x) with respect to x at x = e?
Which of the following equations is/are correct?
1. f(-2) = f(5)
2. f”(-2) + f”(0.5) + f”(3) = 4
Select the correct answer using the code given below:
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What is the derivative of \(\rm e^{e^x}\) with respect to e x?
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