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Chapter 27: Problem 13

Find the derivatives of the given functions. $$y=6 x \tan ^{-1}(1 / x)$$

Short Answer

Expert verified
The derivative is \( 6 \tan^{-1}(1/x) - \frac{6x}{x^2 + 1} \).

Step by step solution

01

Apply Product Rule

The function is a product of two functions: \( u(x) = 6x \) and \( v(x) = \tan^{-1}(1/x) \). We will apply the product rule: \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \). First, differentiate \( u(x) = 6x \), resulting in \( u'(x) = 6 \).
02

Differentiate \( v(x) = \tan^{-1}(1/x) \)

Next, differentiate \( v(x) = \tan^{-1}(1/x) \). Set \( z = 1/x \), making \( v(x) = \tan^{-1}(z) \). The derivative of \( \tan^{-1}(z) \) is \( \frac{1}{1+z^2} \). Now, differentiate \( z = 1/x \) to obtain \( \frac{dz}{dx} = -1/x^2 \). Apply the chain rule to find \( v'(x) = \frac{1}{1+(1/x)^2} \times (-1/x^2) = \frac{-1}{x^2 + 1} \).
03

Substitute into the Product Rule Formula

Insert \( u'(x) = 6 \), \( v(x) = \tan^{-1}(1/x) \), \( u(x) = 6x \), and \( v'(x) = \frac{-1}{x^2 + 1} \) into the product rule formula: \[\frac{d}{dx} [6x \tan^{-1}(1/x)] = 6 \tan^{-1}(1/x) + 6x \left(\frac{-1}{x^2 + 1}\right)]\].
04

Simplify the Expression

Simplify the expression from Step 3: \[ \frac{dy}{dx} = 6 \tan^{-1}(1/x) - \frac{6x}{x^2 + 1} \]. This is the derivative of the function \( y = 6x \tan^{-1}(1/x) \).

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Key Concepts

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

Understanding the Product Rule
When faced with the task of finding the derivative of a product of two functions, the product rule is your go-to tool. Imagine you have two functions, referred to here as \( u(x) \) and \( v(x) \). The product of these functions is \( u(x) \, v(x) \). To find the derivative of this product, the product rule comes into play:
  • The formula is: \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \).
  • This tells us to take the derivative of the first function, \( u'(x) \), and multiply it by the second function, \( v(x) \).
  • Then, add the product of the first function, \( u(x) \), and the derivative of the second function, \( v'(x) \).
Using a simple example, if you have \( u(x) = 6x \) and \( v(x) = \tan^{-1}(1/x) \), first find \( u'(x) \) by differentiating \( 6x \), resulting in \( 6 \).Then, find \( v'(x) \) which involves using the chain rule, as we'll see next.
Applying the Chain Rule
The chain rule is a crucial differentiation technique when dealing with composite functions鈥攆unctions within functions. If you have a function like \( g(f(x)) \), the chain rule helps us differentiate it by working from the inside out.
  • The formula is: \( \frac{d}{dx}[g(f(x))] = g'(f(x)) \cdot f'(x) \).
  • Identify the outer function \( g \) and the inner function \( f \).
  • Differentiate \( g \) with respect to the inner function \( f \), then multiply by \( f' \), the derivative of the inner function.
In the original exercise, the function \( v(x) = \tan^{-1}(1/x) \) can be seen as \( \tan^{-1}(z) \) where \( z = 1/x \). First, find the derivative of \( \tan^{-1}(z) \), which is \( \frac{1}{1+z^2} \). Then, differentiate \( z = 1/x \) to get \( \frac{dz}{dx} = -1/x^2 \). Combining these gives \( v'(x) = \frac{-1}{x^2 + 1} \). This shows how derivatives can untangle even the most intertwined functions.
Inverse Trigonometric Functions Unraveled
Inverse trigonometric functions like \( \tan^{-1}(x) \) can initially seem intimidating when it comes to differentiation. However, by understanding their derivatives, much of the mystery is resolved.
  • The basic derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \).
  • For each inverse trig function, there is a corresponding derivative, often involving fractions that represent rates of change for these functions.
  • Remember, these derivatives often use identities and special forms to simplify and solve equations.
For example, in the problem \( \tan^{-1}(1/x) \) was differentiated using substitution. Recognizing these derivatives allows us to tackle various complex expressions with confidence, applying them correctly within context.
Exploring Differentiation Techniques
Differentiation lies at the heart of calculus, enabling us to understand change and slopes of curves. A variety of techniques are used to handle different forms of functions.
  • Product Rule: As seen, it helps when multiplying two or more functions.
  • Chain Rule: Solves "functions within functions."
  • Inverse Trig Functions: Have their own unique rules for differentiation.
These techniques ensure that regardless of a function鈥檚 complexity, we can systematically break it down to find its derivative. Each technique can be adapted and combined based on what the problem demands, making them versatile tools in any mathematician's toolkit. As you practice, remember that seeing the pattern in functions and predicting the form of their derivatives is a skill that improves over time.

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Most popular questions from this chapter

Solve the given problems. If \(y=A e^{k x}+B e^{-k x},\) show that \(y^{\prime \prime}=k^{2} y\)

Solve the given problems. Display the graphs of \(y_{1}=\cos x\) and \(y_{n}=\frac{\sin (x+h)-\sin x}{h}\) on the same screen of a calculator for \(0 \leq x \leq 2 \pi .\) For \(n=2,\) let \(h=0.5 ;\) for \(n=3,\) let \(h=0.1 .\) (You might try some even smaller values of \(h .\) ) Use a heavier curve for \(y_{3} .\) What do these curves show?

Solve the given problems by finding the appropriate derivative. A connecting rod \(4 \mathrm{cm}\) long connects a piston to a crank \(2 \mathrm{cm}\) in radius (see Fig. 27.44 ). The acceleration of the piston is given by \(a=2 w^{2}(\cos \theta+0.5 \cos 2 \theta) .\) If the angular velocity \(w\) is constant, find \(\theta\) for maximum or minimum \(a\).

Use the following information. The hyperbolic sine of \(u\) is defined as \(\sinh u=\frac{1}{2}\left(e^{u}-e^{-u}\right)\) Figure 27.37 shows the graph of \(y=\sinh x\) The hyperbolic cosine of \(u\) is defined as \(\cosh u=\frac{1}{2}\left(e^{\mu}+\mathrm{e}^{-u}\right)\) Figure 27.38 shows the graph of \(y=\cosh x\) These functions are called hyperbolic functions since, if \(x=\cosh u\) and \(y=\sinh u, x\) and \(y\) satisfy the equation of the hyperbola \(x^{2}-y^{2}=1\) Verify the fact that the exponential expressions for the hyperbolic sine and hyperbolic cosine given above satisfy the equation of the hyperbola.

Find the derivatives of the given functions. $$z=0.2 \cos (4 \sin 3 \phi)$$
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