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Chapter 3: Problem 24

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

Short Answer

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

Step by step solution

01

Differentiate both sides with respect to x

Given the equation: \(x^{n} y^{m}=(x+y)^{n+m}\) Differentiate both sides with respect to \(x\): \(\frac{d}{dx}\left(x^{n} y^{m}\right)=\frac{d}{dx}\left((x+y)^{n+m}\right)\)
02

Apply the product rule on the left-hand side

Using the product rule on \(x^{n} y^{m}\), \(\frac{d}{dx}\left(x^{n} y^{m}\right)=n x^{n-1} y^{m} + x^{n} m y^{m-1} \cdot \frac{dy}{dx}\)
03

Differentiate the right-hand side

Differentiate \((x+y)^{n+m}\) with respect to \(x\): \(\frac{d}{dx}\left((x+y)^{n+m}\right) = (n+m)(x+y)^{n+m-1}\)
04

Set the equations equal

Set the differentiated left-hand side equal to the differentiated right-hand side: \(n x^{n-1} y^{m} + x^{n} m y^{m-1} \frac{dy}{dx} = (n+m)(x+y)^{n+m-1}\)
05

Assume initial condition for simplification

Assume \(n=1, m=1\): \(x y' = x^{2} (x+y)^{-1}\)
06

Solve for \(y'\)

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Let the function \(f\) be defined by $$ f(x)= \begin{cases}\frac{g(x)-g(a)}{x-a} & \text { if } x \neq a \\\ g^{\prime}(a) & \text { if } x=a\end{cases} $$ Prove that if \(g^{\prime}(a)\) exists, \(f\) is continuous at \(a\).

In Exercises 3 through 8 , a particle is moving along a horizontal line according to the given equation of motion, where \(s \mathrm{ft}\) is the directed distance of the particle from a point \(O\) at \(t \mathrm{sec}\). Find the instantaneous velocity \(v\left(t_{1}\right) \mathrm{ft} / \mathrm{sec}\) at \(t_{1} \mathrm{sec}\); and then find \(v\left(t_{1}\right)\) for the particular value of \(t_{1}\) given. $$ s=3 t^{2}+1 ; t_{1}=3 $$

If a ball is given a push so that it has an initial velocity of \(24 \mathrm{ft} / \mathrm{sec}\) down a certain inclined plane, then \(s=24 t+10 t^{2}\), where \(s \mathrm{ft}\) is the distance of the ball from the starting point at \(t \mathrm{sec}\) and the positive direction is down the inclined plane. (a) What is the instantaneous velocity of the ball at \(t_{1}\) sec? (b) How long does it take for the velocity to increase, to \(48 \mathrm{ft} / \mathrm{sec} ?\)

Do each of the following: (a) Draw a sketch of the graph of the function; (b) determine if \(f\) is continuous at \(x_{1} ;\) (c) find \(f^{\prime}-\left(x_{1}\right)\) and \(f_{+}^{\prime}\left(x_{1}\right)\) if they exist; (d) determine if \(f\) is differentiable at \(x_{1}\). $$ f(x)=(x-2)^{-2} ; x_{1}=2 $$

If an object falls from rest, its equation of motion is \(s=-16 t^{2}\), where \(t\) is the number of seconds in the time that has elapsed since the object left the starting point, \(s\) is the number of feet in the distance of the object from the starting point at \(t \mathrm{sec}\), and the positive direction is upward. If a stone is dropped from a building \(256 \mathrm{ft}\) high, find: (a) the instantaneous velocity of the stone 1 sec after it is dropped; (b) the instantaneous velocity of the stone 2 sec after it is dropped; (c) how long it takes the stone to reach the ground; (d) the instantaneous velocity of the stone when it reaches the ground.
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