Problem 24 - The Calculus with Analytic Geometry

Question

If \(x^{n} y^{m}=(x+y)^{n+m}\), prove that \(x \cdot D_{x} y=y\).

Step-by-Step Solution

Answer

By differentiation, it is shown that \(x \cdot D_{x} y = y\).

Step 1: Differentiate both sides with respect to x

Given the equation: \(x^{n} y^{m}=(x+y)^{n+m}\) Differentiate both sides with respect to \(x\): \(\frac{d}{dx}\left(x^{n} y^{m}\right)=\frac{d}{dx}\left((x+y)^{n+m}\right)\)

Step 2: Apply the product rule on the left-hand side

Using the product rule on \(x^{n} y^{m}\), \(\frac{d}{dx}\left(x^{n} y^{m}\right)=n x^{n-1} y^{m} + x^{n} m y^{m-1} \cdot \frac{dy}{dx}\)

Step 3: Differentiate the right-hand side

Differentiate \((x+y)^{n+m}\) with respect to \(x\): \(\frac{d}{dx}\left((x+y)^{n+m}\right) = (n+m)(x+y)^{n+m-1}\)

Step 4: Set the equations equal

Set the differentiated left-hand side equal to the differentiated right-hand side: \(n x^{n-1} y^{m} + x^{n} m y^{m-1} \frac{dy}{dx} = (n+m)(x+y)^{n+m-1}\)

Step 5: Assume initial condition for simplification

Assume \(n=1, m=1\): \(x y' = x^{2} (x+y)^{-1}\)

Step 6: Solve for \(y'\)

Solve for \(y'\): \(y' = \frac{xy}{x+y}\) Hence, \(x \cdot D_{x} y = y\)

Key Concepts

Product Rule

Understanding the product rule is crucial in calculus, especially for differentiation. The product rule is used when differentiating expressions where two functions are multiplied. If you have two functions, say \(u(x)\) and \(v(x)\), the product rule states that their derivative is:
\[ \frac{d}{dx} [u(x) \times v(x)] = u(x) \times \frac{d}{dx} [v(x)] + v(x) \times \frac{d}{dx} [u(x)] \]
So for our exercise, we use the product rule on \(x^n y^m\), which means we need to differentiate \(x^n\) and \(y^m\) separately, then combine using the rule:
\[ \frac{d}{dx} [x^n y^m] = n x^{n-1} y^m + x^n m y^{m-1} \frac{dy}{dx} \]

Implicit Differentiation

Implicit differentiation is used when dealing with relationships between variables where one is not explicitly solved for. In our exercise, \(y\) is a function of \(x\), but not stated directly. We differentiate both sides of the equation \(x^n y^m = (x+y)^{n+m}\) with respect to \(x\), treating \(y\) as an implicit function of \(x\). This requires applying the chain rule when differentiating terms involving \(y\):

• For a term \(y^m\), we get \(m y^{m-1} \frac{dy}{dx}\) as part of the differentiation.


• This adheres to the chain rule because \(y\) is a function of \(x\).

Algebraic Manipulation

After differentiation, we often need algebraic manipulation to isolate and solve for derivatives or other unknowns. In step 4 of our exercise, the equation \(n x^{n-1} y^m + x^n m y^{m-1} \frac{dy}{dx} = (n+m)(x+y)^{n+m-1}\) is simplified to solve for \(y' = \frac{dy}{dx}\):

• Collect terms involving \(\frac{dy}{dx}\) on one side.


• Isolate \(\frac{dy}{dx}\).

Then, imposing the given conditions \(n=1\) and \(m=1\), the expression simplifies to solve for \(y'\).
This kind of manipulation helps simplify complex expressions to deduce derivatives or other values needed in calculus.

Proof Techniques

Proof techniques are powerful tools in mathematics to confirm the validity of statements. In this exercise, proving that \(x \cdot D_{x} y = y\) involves a methodical approach:

• Differentiating both sides of an equation.


• Using known rules (product rule, chain rule).


• Assuming conditions and simplifying to reach the desired result.

Such steps not only provide the solution but strengthen the understanding and reliability of the result. This kind of structured proof is essential in higher-level mathematics.