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Chapter 20: Problem 3

If \(y\) is a function of \(x\), and \(x=\frac{e^{t}}{e^{t}+1}\) show that \(\frac{\mathrm{d} y}{\mathrm{~d} t}=x(1-x) \frac{\mathrm{d} y}{\mathrm{~d} x}\).

Short Answer

Expert verified
\[\frac{dy}{dt} = x(1-x)\frac{dy}{dx}\]

Step by step solution

01

Differentiating x with respect to t

To find \(\frac{dx}{dt}\), we will differentiate \(x = \frac{e^t}{e^t + 1}\) with respect to \(t\). Using the quotient rule, we have: \[\frac{dx}{dt} = \frac{(e^t + 1)(e^t) - e^t (e^t)}{(e^t + 1)^2}\]
02

Simplify the expression

Now, we will simplify the obtained expression: \[\frac{dx}{dt} = \frac{e^{2t} + e^t - e^{2t}}{(e^t + 1)^2} = \frac{e^t}{(e^t + 1)^2}\]
03

Applying the Chain Rule

Now, we will apply the Chain Rule. Recall that it states that \(\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}\). We want to find the relationship between \(\frac{dy}{dt}\) and \(\frac{dy}{dx}\). We have already found \(\frac{dx}{dt}\), so the equation becomes: \[\frac{dy}{dt} = \frac{dy}{dx}\frac{e^t}{(e^t + 1)^2}\]
04

Expressing \(\frac{dy}{dt}\) in terms of x

Our goal is to rewrite this equation to express \(\frac{dy}{dt}\) in terms of \(x\). Recall that \(x = \frac{e^t}{e^t + 1}\). We will replace \(e^t\) in the numerator with \(x(e^t + 1)\) as follows: \[\frac{dy}{dt} = \frac{dy}{dx}\frac{x(e^t + 1)}{(e^t + 1)^2}\] Canceling out a factor of \((e^t + 1)\) from the numerator and denominator, we get: \[\frac{dy}{dt} = \frac{dy}{dx}\frac{x}{e^t + 1}\] Again, look at the expression \(x = \frac{e^t}{e^t + 1}\). By multiplying both sides of this equation by \(e^t + 1\), we get: \[x(e^t + 1) = e^t\] But since \(x = \frac{e^t}{e^t + 1}\), it follows that \(1 - x = \frac{1}{e^t + 1}\). Thus, we can substitute this into the expression for \(\frac{dy}{dt}\): \[\frac{dy}{dt} = \frac{dy}{dx}(1-x)\]
05

Conclusion

Now, we have the relationship between \(\frac{dy}{dt}\) and \(\frac{dy}{dx}\) in a simple form: \[\frac{dy}{dt} = x(1-x)\frac{dy}{dx}\] This is the desired result.

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Key Concepts

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

Chain Rule
The Chain Rule in calculus is a fundamental technique for finding the derivative of a composition of functions. It's particularly useful when dealing with complex functions where one function is nested inside another. Suppose you have two functions, say \( f \) and \( g \), and you want to differentiate the composition \( f(g(x)) \). The Chain Rule tells us that the derivative is given by:
  • \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \).
In practical terms, this means you first differentiate the outer function \( f \) with respect to its argument \( g(x) \) and multiply by the derivative of the inner function \( g \) with respect to \( x \). The Chain Rule is essential for differentiating implicit functions and functions of a variable, as it allows the transformation of derivatives between variables.
Derivatives
Derivatives are a central concept in differential calculus, describing the rate at which a function changes. Mathematically, the derivative of a function \( y=f(x) \) at a point \( x \) is given by:
  • \( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \).
This limit captures the instantaneous rate of change or the slope of the tangent to the curve defined by the function at a specific point.
Derivatives can also be viewed as the linear approximation of a function at a given point or the function's sensitivity to change in its input variable. They are used extensively in finding maxima and minima of functions, analyzing function behavior, and solving real-world problems, such as calculating velocity in physics.
Functions of a Variable
Functions of a variable are mathematical entities that describe relationships between dependent and independent variables. For a function \( y = f(x) \), \( x \) is often called the independent variable, while \( y \) is the dependent variable since its value depends on the chosen \( x \).
An expression like \( x=\frac{e^{t}}{e^{t}+1} \), where \( x \) is dependent on another variable \( t \), introduces an additional layer of dependency. This is often encountered in real-life situations where one quantity depends on another, such as population growth over time.
  • The notion of chain rule arises when dealing with such relationships as it lets us connect the rates of change (derivatives) of different variables effectively.
  • By understanding how one variable affects another, we are often able to derive more complex relationships and solve intricate problems efficiently.
By exploring functions of a variable, we can understand more about the interconnectivity of dynamic systems and conduct better mathematical analyses.

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Most popular questions from this chapter

$$ \text { If } x=3 \cos \theta-\cos ^{3} \theta, y=3 \sin \theta-\sin ^{3} \theta \text {, express } \frac{\mathrm{d} y}{\mathrm{~d} x} \text { and } \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} \text { in terms of } \theta \text {. } $$

Write out the solutions carefully. They are all quite straightforward. Find an expression for \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) when \(x^{3}+y^{3}+4 x y^{2}=5\).

Write out the solutions carefully. They are all quite straightforward. If \(x=3(1-\cos \theta)\) and \(y=3(\theta-\sin \theta)\) find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) and \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\) in their simplest forms.

If \(y=e^{-k t}(A \cosh q t+B \sinh q t)\) where \(A, B, q\) and \(k\) are constants, show that \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}}+2 k \frac{\mathrm{d} y}{\mathrm{~d} t}+\left(k^{2}-q^{2}\right) y=0 .\)

Show that \(y=e^{-2 m x} \sin 4 m x\) is a solution of the equation \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+4 m \frac{\mathrm{d} y}{\mathrm{~d} x}+20 m^{2} y=0\)
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