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Chapter 2: Problem 72

Find the derivative of the function. $$ f(x)=x \tan x \sec ^{-1} x $$

Short Answer

Expert verified
The derivative of the function \(f(x) = x \tan{x} \sec^{-1}{x}\) is: \(f'(x) = \tan{x} \sec^{-1}{x} + x\sec^2{x} \sec^{-1}{x} + \tan{x}\left(\frac{1}{\sqrt{1-x^2}}\right)\).

Step by step solution

01

Identify the functions

In this exercise, we have three functions: f(x) = x, g(x) = tan(x), and h(x) = sec^(-1)(x).
02

Find the derivatives of the functions

First, we need to find the derivatives of f(x), g(x), and h(x). The derivative of f(x) = x is f'(x) = 1. For g(x) = tan(x), recall that the derivative of tan(x) is sec^2(x), so we have g'(x) = sec^2(x). For h(x) = sec^(-1)(x), recall that the derivative of sec^(-1)(x) is \[\frac{1}{x\sqrt{1-x^2}}\]. Therefore, h'(x) = \[\frac{1}{x\sqrt{1-x^2}}\].
03

Apply the product rule

Now let's apply the product rule to find the derivative of f(x) * g(x) * h(x). We will do this in two rounds: first, apply the product rule for f(x) * g(x) = w(x) and then for w(x) * h(x). Product rule for f(x) * g(x): w'(x) = f'(x) * g(x) + f(x) * g'(x) w'(x) = 1 * tan(x) + x * sec^2(x) Now apply the product rule for w(x) * h(x): (w(x) * h(x))' = w'(x) * h(x) + w(x) * h'(x)
04

Compute the derivative

Using the derivatives we found earlier, we get: \( (w(x) * h(x))' = (1 * tan(x) + x * sec^2(x)) * sec^(-1)(x) + (x * tan(x)) * (\frac{1}{x\sqrt{1-x^2}}) \) Simplifying the expression: \( (w(x) * h(x))' = tan(x) * sec^(-1)(x) + x * sec^2(x) * sec^(-1)(x) + tan(x) * (\frac{1}{\sqrt{1-x^2}}) \) So, the derivative of f(x) = x * tan(x) * sec^(-1)(x) is: \( f'(x) = tan(x) * sec^(-1)(x) + x * sec^2(x) * sec^(-1)(x) + tan(x) * (\frac{1}{\sqrt{1-x^2}}) \)

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Key Concepts

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

Product Rule in Differentiation
Understanding the product rule is vital when dealing with the differentiation of functions that are products of two or more functions. The product rule states that if you have two functions, let's call them u(x) and v(x), the derivative of their product u(x)v(x) is given by u'(x)v(x) + u(x)v'(x), where u'(x) is the derivative of u(x) and v'(x) is the derivative of v(x).

For example, when confronted with the function f(x) = x * tan(x) * sec-1(x), we must first consider the product of x and tan(x) before we can incorporate the third function, sec-1(x), using the product rule. By applying the rule systematically, the complex expression is broken down into manageable parts, allowing for accurate differentiation step by step.
  • First, identify u(x) as the first function and v(x) as the second.
  • Then, find their individual derivatives u'(x) and v'((x).
  • Lastly, apply the product rule to find the derivative of their product.
Remember, when the problem involves more than two functions, you can extend the product rule: for functions u(x), v(x), and w(x), the derivative of their product is found by taking the derivative of the product of the first two (using the product rule) and then using the product rule again with the third function, as seen in our example.
Chain Rule for Composite Functions
The chain rule is a key concept in calculus used to find the derivative of composite functions. Essentially, if you have a function h(x) which is composed of another function g(f(x)), to find h'(x), you need to multiply the derivative of the outer function g'(f(x)) by the derivative of the inner function f'(x). This rule reflects how rates of change in functions affect each other.

For trigonometric functions, the chain rule becomes particularly useful when these functions are composed with other functions, such as a polynomial or another trigonometric function. It can be daunting at first, but with practice, applying the chain rule becomes a straightforward, mechanical process.
  • First, identify the outer and inner functions in the composition.
  • Calculate their individual derivatives.
  • Finally, multiply these derivatives according to the chain rule.
Using the chain rule helps unpack the derivatives of more complex expressions, and mastering it can greatly simplify solving calculus problems.
Implicit Differentiation and Its Applications
Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. In other words, both variables are intermingled on one side of the equation. This method is particularly helpful when dealing with equations that are difficult or impossible to solve for one variable.

For implicit differentiation, follow the next steps:
  • Differentiate both sides of the equation with respect to x, treating y as a function of x (if y is the other variable involved).
  • Apply the differentiation rules like the product rule, chain rule, or quotient rule where necessary.
  • Finally, solve for y' (dy/dx) to find the derivative.
Implicit differentiation enables the handling of complex functions and relations that are not in the 'y equals' format, broadening the scope of problems that can be tackled in calculus. It's a powerful tool that aids in understanding how related variables change with respect to each other in intricate mathematical relationships.

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Most popular questions from this chapter

Electric Potential Suppose that a ring-shaped conductor of radius \(a\) carries a total charge \(Q\). Then the electrical potential at the point \(P\), a distance \(x\) from the center and along the line perpendicular to the plane of the ring through its center, is given by $$ V(x)=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{\sqrt{x^{2}+a^{2}}} $$ where \(\varepsilon_{0}\) is a constant called the permittivity of free space. The magnitude of the electric field induced by the charge at the point \(P\) is \(E=-d V / d x\), and the direction of the field is along the \(x\) -axis. Find \(E\).

Two curves are said to be orthogonal if their tangent lines are perpendicular at each point of intersection of the curves. In Exercises \(89-92\), show that the curves with the given equations are orthogonal. $$ x^{2}+3 y^{2}=4, \quad y=x^{3} $$

Continuous Compound Interest Formula See Section 3.5. Use l'H么pital's Rule to derive the continuous compound interest formula $$ A=P e^{r t} $$ where \(A\) is the accumulated amount, \(P\) is the principal, \(t\) is the time in years, and \(r\) is the nominal interest rate per year compounded continuously, from the compound interest formula $$ A=P\left(1+\frac{r}{m}\right)^{m t} $$ where \(r\) is the nominal interest rate per year compounded \(m\) times per year.

If a right circular cylinder of radius \(a\) is filled with water and rotated about its vertical axis with a constant angular velocity \(\omega\), then the water surface assumes a shape whose cross section in a plane containing the vertical axis is a parabola. If we choose the \(x y\) -system so that the \(y\) -axis is the axis of rotation and the vertex of the parabola passes through the origin of the coordinate system, then the equation of the parabola is $$y=\frac{\omega^{2} x^{2}}{2 g}$$ where \(g\) is the acceleration due to gravity. Find the angle \(\alpha\) that the tangent line to the water level makes with the \(x\) -axis at any point on the water level. What happens to \(\alpha\) as \(\omega\) increases? Interpret your result.

Potential of a Charged Disk The potential on the axis of a uniformly charged disk is $$ V(r)=\frac{\sigma}{2 \varepsilon_{0}}\left(\sqrt{r^{2}+R^{2}}-r\right) $$ where \(\varepsilon_{0}\) and \(\sigma\) are constants. The force corresponding to this potential is \(F(r)=-V^{\prime}(r) .\) Find \(F(r)\).
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