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Chapter 3: Problem 71

Use a trigonometric identity to show that the derivatives of the inverse cotangent and inverse cosecant differ from the derivatives of the inverse tangent and inverse secant, respectively, by a multiplicative factor of -1

Short Answer

Expert verified
Question: Show that the derivatives of the inverse cotangent (arccot) and inverse cosecant (arccsc) differ from the derivatives of the inverse tangent (arctan) and inverse secant (arcsec), respectively, by a multiplicative factor of -1. Answer: The derivatives of the inverse cotangent (arccot) and inverse cosecant (arccsc) are -1/(1 + x^2) and -1/(|x| * sqrt(x^2 - 1)), respectively. When compared to the derivatives of the inverse tangent (arctan) and inverse secant (arcsec), which are 1/(1 + x^2) and 1/(|x| * sqrt(x^2 - 1)), respectively, it can be seen that the derivatives of arccot and arccsc differ from those of arctan and arcsec by a multiplicative factor of -1.

Step by step solution

01

Find the derivative of arccot(x) and arccsc(x)

To find the derivative of arccot(x) and arccsc(x), we will first use the following identities: - arccot(x) = arctan(1/x) - arccsc(x) = arcsec(1/x) Now, we will apply the chain rule on both identities: For arccot(x): (dy/dx)[arctan(1/x)] = (dy/du)[arctan(u)] * (du/dx)[1/x], where u = 1/x For arccsc(x): (dy/dx)[arcsec(1/x)] = (dy/du)[arcsec(u)] * (du/dx)[1/x], where u = 1/x By differentiating the inverse trigonometric functions, we have: (dy/du)[arctan(u)] = 1/(1 + u^2) (dy/du)[arcsec(u)] = 1/(|u| * sqrt(u^2 - 1)) Computing the derivatives, we get: Derivative of arccot(x): (dy/dx)[arctan(1/x)] = -1/(1 + x^2) Derivative of arccsc(x): (dy/dx)[arcsec(1/x)] = -1/(|x| * sqrt(x^2 - 1))
02

Show that the derivatives differ by a factor of -1

Now, we will compare the derivatives of arccot(x) and arccsc(x) with the derivatives of arctan(x) and arcsec(x), respectively. Derivative of arctan(x): (dy/dx)[arctan(x)] = 1/(1 + x^2) Derivative of arcsec(x): (dy/dx)[arcsec(x)] = 1/(|x| * sqrt(x^2 - 1)) Comparing the derivatives, we can see that: Derivative of arccot(x) = -1 * Derivative of arctan(x) Derivative of arccsc(x) = -1 * Derivative of arcsec(x) Hence, the derivatives of the inverse cotangent and inverse cosecant differ from the derivatives of the inverse tangent and inverse secant, respectively, by a multiplicative factor of -1.

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