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Chapter 11: Problem 2

Use the quotient rule to find the derivatives of the following: (a) \(\frac{\cos x}{\sin x}\) (b) \(\frac{\tan t}{\ln t}\) (c) \(\frac{\mathrm{e}^{2 t}}{t^{3}+1}\) (d) \(\frac{3 x^{2}+2 x-9}{x^{3}+1}\) (e) \(\frac{x^{2}+x+1}{1+e^{x}}\) (f) \(\frac{\sinh 2 t}{\cosh 3 t}\) (g) \(\frac{1+\mathrm{e}^{t}}{1+\mathrm{e}^{2 t}}\)

Short Answer

Expert verified
a) \(-\csc^2 x\); b) \(\frac{\ln t \cdot \sec^2 t - \tan t/t}{(\ln t)^2}\); c) \(\frac{2\mathrm{e}^{2t}(t^3+1) - 3t^2\mathrm{e}^{2t}}{(t^3+1)^2}\); d) \(\frac{(x^3+1)(6x+2)-(3x^2+2x-9)(3x^2)}{(x^3+1)^2}\); e) \(\frac{(1+e^x)(2x+1)-(x^2+x+1)e^x}{(1+e^x)^2}\); f) \(\frac{\cosh 3t \cdot 2\cosh 2t - \sinh 2t \cdot 3\sinh 3t}{(\cosh 3t)^2}\); g) \(\frac{(1+e^{2t})e^t-(1+e^t)2e^{2t}}{(1+e^{2t})^2}\).

Step by step solution

01

Understanding the Quotient Rule

The quotient rule is a formula used to find the derivative of a quotient of two functions. If we have two functions, \( u(x) \) and \( v(x) \), the derivative of their quotient \( \frac{u}{v} \) is given by \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \). Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) with respect to \( x \).
02

Exercise (a): Differentiating \( \frac{\cos x}{\sin x} \)

Identify \( u = \cos x \) and \( v = \sin x \). Their derivatives are \( u' = -\sin x \) and \( v' = \cos x \). Applying the quotient rule, \( \frac{v \cdot u' - u \cdot v'}{v^2} = \frac{\sin x \cdot (-\sin x) - \cos x \cdot \cos x}{(\sin x)^2} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x} = -\csc^2 x \).
03

Exercise (b): Differentiating \( \frac{\tan t}{\ln t} \)

Let \( u = \tan t \) and \( v = \ln t \). Then \( u' = \sec^2 t \) and \( v' = \frac{1}{t} \). Using the quotient rule, \( \frac{\ln t \cdot \sec^2 t - \tan t \cdot \frac{1}{t}}{(\ln t)^2} = \frac{\ln t \cdot \sec^2 t - \frac{\tan t}{t}}{(\ln t)^2} \).
04

Exercise (c): Differentiating \( \frac{\mathrm{e}^{2t}}{t^3+1} \)

Here, \( u = \mathrm{e}^{2t} \) and \( v = t^3 + 1 \). So, \( u' = 2\mathrm{e}^{2t} \) and \( v' = 3t^2 \). Applying the quotient rule gives \( \frac{(t^3+1) \cdot 2\mathrm{e}^{2t} - \mathrm{e}^{2t} \cdot 3t^2}{(t^3+1)^2} = \frac{2\mathrm{e}^{2t}(t^3+1) - 3t^2\mathrm{e}^{2t}}{(t^3+1)^2} \).
05

Exercise (d): Differentiating \( \frac{3x^2 + 2x - 9}{x^3 + 1} \)

Let \( u = 3x^2 + 2x - 9 \) and \( v = x^3 + 1 \). Then, \( u' = 6x + 2 \) and \( v' = 3x^2 \). Using the quotient rule, \( \frac{(x^3 + 1)(6x + 2) - (3x^2 + 2x - 9)(3x^2)}{(x^3 + 1)^2} \).
06

Exercise (e): Differentiating \( \frac{x^2 + x + 1}{1 + e^x} \)

Here, \( u = x^2 + x + 1 \) and \( v = 1 + e^x \). Thus, \( u' = 2x + 1 \) and \( v' = e^x \). The quotient rule gives \( \frac{(1 + e^x)(2x + 1) - (x^2 + x + 1)e^x}{(1 + e^x)^2} \).
07

Exercise (f): Differentiating \( \frac{\sinh 2t}{\cosh 3t} \)

Define \( u = \sinh 2t \) and \( v = \cosh 3t \). Therefore, \( u' = 2\cosh 2t \) and \( v' = 3\sinh 3t \). Applying the rule, we have \( \frac{\cosh 3t \cdot 2\cosh 2t - \sinh 2t \cdot 3\sinh 3t}{(\cosh 3t)^2} \).
08

Exercise (g): Differentiating \( \frac{1+\mathrm{e}^{t}}{1+\mathrm{e}^{2t}} \)

Set \( u = 1 + \mathrm{e}^{t} \) and \( v = 1 + \mathrm{e}^{2t} \). So, \( u' = \mathrm{e}^{t} \) and \( v' = 2\mathrm{e}^{2t} \). Using the quotient rule, \( \frac{(1 + \mathrm{e}^{2t})\mathrm{e}^{t} - (1 + \mathrm{e}^{t})2\mathrm{e}^{2t}}{(1 + \mathrm{e}^{2t})^2} \).

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Key Concepts

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

Differentiation
Differentiation is a key concept in calculus that involves finding the rate at which a function changes at any given point. It helps understand how a function behaves and is particularly useful in identifying the local maxima and minima of functions. Differentiation is foundational for solving real-world problems involving rates of change, such as speed, acceleration, and other variables that involve time.
  • It uses the derivative of a function, which provides the best linear approximation of the function at any specific point.
  • The process of finding a derivative is often symbolically represented as \( dy/dx \) for a function \( y = f(x) \).
  • To differentiate a function means determining its gradient or slope at any point, which is crucial for understanding its behavior.
When working with complex functions, formulas like the product rule, quotient rule, and chain rule simplify the process of differentiation. Each rule has its own specific application depending on the nature of the function involved.
Derivatives
Derivatives are the core outcome of the differentiation process. They express the rate of change of a quantity and are fundamental in the study of calculus. The derivative of a function describes how the function's output value changes as the input value changes.
  • The derivative is often represented by \( f'(x) \) or \( \frac{dy}{dx} \), denoting how a change in \( x \) affects \( y \).
  • In practical terms, a derivative can represent quantities like velocity, which is the rate of change of distance with respect to time.
Understanding derivatives helps analysts predict future behavior and patterns of curves. Higher order derivatives are also used to explore the curvature and concavity of functions, giving a deeper insight into the dynamics at play.
Calculus
Calculus is a branch of mathematics that studies continuous change, much like algebra studies operations and their application to solving equations. Calculus is divided into differential calculus and integral calculus, each focusing on different problems but interconnected as fundamental aspects of the study of change.
  • Differential Calculus: Primarily concerned with the notion of a derivative, it deals with how functions change and the concept of slope and rates of change.
  • Integral Calculus: Focuses on the accumulation of quantities and the areas under and between curves. It deals with the inverse process of differentiation.
The concepts of calculus are indispensable in modern science and engineering for modeling dynamic systems. It provides tools to describe the motion of planets, analyze the fluid dynamics, and even predict probabilistic events.
Mathematical Functions
In mathematics, a function is a relation that uniquely associates elements of one set with elements of another set. Often expressed as \( f(x) \), functions are used to denote equations, mappings, or real-world processes.
  • A function assigns exactly one output to each input, which is crucial for determining consistent relationships in mathematical models.
  • Mathematical functions come in various forms, such as linear, quadratic, polynomial, trigonometric, exponential, and logarithmic functions.
Understanding functions and their behavior is crucial in applying calculus for real-world problems. Functions describe anything from the simple oscillation of a pendulum to the more complex growth curves of biological organisms, highlighting their versatility and importance in scientific inquiry.

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Most popular questions from this chapter

Differentiate (a) \(y=\ln x\) (b) \(y=\ln 2 x\) (c) \(y=\ln k x, k\) constant (d) \(y=\ln (1+t)\) (e) \(y=\ln (3+4 t)\) (f) \(y=\ln (5+7 \sin x)\)

\text { If } x=t+t^{2}+t^{3} \text { and } y=\sin 2 t \text {, find } \frac{\mathrm{d} y}{\mathrm{~d} x} \text { when } t=1 \text {. }

Use the product rule to differentiate the following functions: (a) \(y=\sin x \cos x\) (b) \(y=\ln t \tan t\) (c) \(y=\left(t^{3}+1\right) \mathrm{e}^{2 t}\) (d) \(y=\sqrt{x} \mathrm{e}^{x}\) (e) \(y=\mathrm{e}^{t} \sin t \cos t\) (f) \(y=3 \sinh 2 t \cosh 3 t\) (g) \(y=(1+\sin t) \tan t\) (h) \(y=4 \sinh (t+1) \cosh (1-t)\)

Find the equation of the tangent to $$ y(x)=\mathrm{e}^{3 x}(1-x) \quad \text { at the point }(0,1) $$

For which values of \(t\) is the derivative of \(y(t)=\mathrm{e}^{-t} t^{2}\) zero?
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