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Chapter 21: Problem 5

Differentiate the function given. \(y=x \tan ^{-1} x\)

Short Answer

Expert verified
The derivative of the given function \(y=x \tan ^{-1} x\) is \(y'= \tan ^{-1} x + \frac{x}{1+x^2}\)

Step by step solution

01

Identify the two functions

In this case, the two functions that are being multiplied together are \(f(x) = x\) and \(g(x) = \tan ^{-1} x\).
02

Find the derivative of each function

Firstly, the derivative of \(f(x) = x\) is \(f'(x) = 1\). The function \(g(x) = \tan ^{-1} x\) might seem more tricky, but you can remember that the derivative of \(arctan(x)\) is \(g'(x) = \frac{1}{1+x^2}\).
03

Apply the product rule

The product rule states that(in this case) \(y' = f'(x)g(x) + f(x)g'(x)\). So, using the derivative of each function as calculated in step 2, derivative of the given function would be \(y'= 1 * \tan ^{-1} x + x * \frac{1}{1+x^2}\). Hence, \(y'= \tan ^{-1} x + \frac{x}{1+x^2}\).

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Key Concepts

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

Product Rule
To differentiate a function that is the product of two simpler functions, we use the product rule. This rule is essential when handling derivatives involving multiplication. Here鈥檚 how it works: the derivative of a product of two functions, say \( u(x) \) and \( v(x) \), is determined by the formula \( (uv)' = u'v + uv' \). In simple terms, this means you differentiate the first function and multiply it by the second function, then add it to the product of the first function and the derivative of the second. For example:
  • Let \( u(x) = x \) and \( v(x) = \tan ^{-1} x \).

  • Then, their derivatives are \( u'(x) = 1 \) and \( v'(x) = \frac{1}{1+x^2} \).

  • Thus, applying the product rule gives \( y' = 1 * \tan ^{-1} x + x * \frac{1}{1+x^2} \).
This formula is very effective for the product of two differentiable functions, ensuring you get the correct derivative every time.
Inverse Trigonometric Functions
Inverse trigonometric functions can often seem challenging, but once you understand their behavior and derivatives, they become much easier to work with. These functions are the inverses of the standard trigonometric functions, like sine, cosine, and tangent. The inverse tangent function, denoted as \( \tan^{-1}(x) \) or arctan \( (x) \), is the function we deal with in this exercise.
  • The derivative of \( \tan^{-1}(x) \) is key for our calculations. It is \( \frac{1}{1+x^2} \).

  • This result forms the backbone when applying the product rule in differentiation where inverse trig functions are involved.
Understanding how to differentiate inverse trigonometric functions allows you to tackle a variety of problems in calculus.
Derivative Calculation
Combining your knowledge of the product rule and inverse trigonometric functions, you can effectively calculate the derivative of more complex expressions. Derivative calculation is the process of finding the rate at which a function changes at any given point. Here's a simple breakdown in the context of our exercise:
  • First, identify the component functions in the product (\( f(x) = x \) and \( g(x) = \tan^{-1}(x) \)).

  • Next, find the derivatives of these components \( (f'(x) = 1, g'(x) = \frac{1}{1+x^2}) \).

  • Apply the product rule to these derivatives, as previously mentioned, to find the derivative of the original function: \( y' = \tan ^{-1} x + \frac{x}{1+x^2} \).
This methodical approach is crucial for solving derivative problems accurately, offering a structured pathway from identifying basic components to arriving at the final derivative.

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Most popular questions from this chapter

Approximating the function \(f(x)=\sin x\) near \(x=0\) by using polynomials. The point of this problem is to show you how the values of \(\sin x\) can be approximated numerically with a very high degree of accuracy. It is an introduction to Taylor polynomials. (a) Find the equation of the line tangent to \(f(x)=\sin x\) at \(x=0\). (b) Find the equation of a quadratic \(Q(x)=a+b x+c x^{2}\) such that the function \(Q(x)\) and its nonzero derivatives match those of \(\sin x\) at \(x=0\). In other words, \(Q(0)=f(0), Q^{\prime}(0)=f^{\prime}(0)\), and \(Q^{\prime \prime}(0)=f^{\prime \prime}(0)\). The quadratic that you found is the quadratic that best "fits" the sine curve near \(x=0\). In fact, the "quadratic" turns out not to really be a quadratic at all. Sine is an odd function, so there is no parabola that "fits" the sine curve well at \(x=0\). (c) Find the equation of a cubic \(C(x)=a+b x+c x^{2}+d x^{3}\) such that the function \(C(x)\) and its nonzero derivatives match those of \(\sin x\) at \(x=0 .\) In other words, \(C(0)=f(0), C^{\prime}(0)=f^{\prime}(0), C^{\prime \prime}(0)=f^{\prime \prime}(0)\), and \(C^{\prime \prime \prime}(0)=f^{\prime \prime \prime}(0) .\) The cubic that you found is the cubic that best "fits" the sine curve near \(x=0\). Using a calculator, on the same set of axes graph \(\sin x\), the tangent line to \(\sin x\) at \(x=0\), and \(C(x)\), the cubic you found. Now "zoom in" around \(x=0 .\) Can you see that near \(x=0\) the line is a good fit to the sine curve but the cubic is an even better fit for small \(x\) and the cubic hugs the sine curve for longer? The next set of questions asks you to investigate how good the fit is. (d) Use \(C(x)\) from part (c) to estimate the following, and then compare with the actual value using a calculator. \(\begin{array}{lllll}\sin (0.01) & \sin (0.1) & \sin (0.5) & \sin (1) & \sin (3)\end{array}\) (e) (Challenge) Find the "best" fifth degree polynomial approximation of \(\sin (x)\) for \(x\) near 0 by making sure that the first five derivatives of the polynomial match those of \(\sin (x)\) when evaluated at \(x=0 .\) Graph \(\sin x\) along with its first, third, and fifth degree polynomial approximations on your graphing calculator. The higher the degree of the polynomial, the better the fit to \(\sin x\) near \(x=0\), right? (f) Using a calculator, on the same set of axes graph \(\sin x\) and the polynomial given below. $$x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\frac{x^{7}}{5040}+\frac{x^{9}}{362880}$$ This polynomial is an even better fit than the last one, right? Now graph the difference between \(\sin x\) and this polynomial; in other words, graph $$y=\sin x-\left[x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\frac{x^{7}}{5040}+\frac{x^{9}}{362880}\right]$$ On approximately what interval is the difference between \(\sin x\) and this polynomial less than \(0.005\) ?

Let \(f(x)=-\cos x\) and \(g(x)=\sin x\). (a) What is the maximum distance between these two curves on the interval \(\left[-\frac{\pi}{4}, \frac{3 \pi}{4}\right] ?\) (b) What is the point of intersection of the tangent lines to these curves at the points from part (a) where the curves are farthest apart? Does this answer surprise you? Explain.

Differentiate \(f(x)=3 \cos \left(\frac{1}{x^{2}+1}\right)+x \arctan \left(\frac{1}{x}\right)\)

Using the derivatives of sine and cosine and either the Product Rule or the Quotient Rule, show that \(\frac{d}{d x} \tan x=\sec ^{2} x\).

A lookout tower is located \(0.5\) kilometers from a line of warehouses. A searchlight on the tower is rotating at a rate of 6 revolutions per minute. How fast is the beam of light moving along the wall of warehouses when it passes by a window located 1 kilometer from the tower?
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