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Find the derivative of the function. h'(x) = = h(x) = (arccos(x)) In(x)

Find the derivative of the function. h'(x) = = h(x) = (arccos(x)) In(x)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem Statement:** Find the derivative of the function. Given: \[ h(x) = (\arccos(x)) \ln(x) \] **Solution:** To find the derivative \( h'(x) \), we will use the product rule, which states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product is given by: \[ (uv)' = u'v + uv' \] In this case, let: \[ u(x) = \arccos(x) \] \[ v(x) = \ln(x) \] First, find the derivatives of \( u(x) \) and \( v(x) \): - The derivative of \( u(x) = \arccos(x) \) is: \[ u'(x) = -\frac{1}{\sqrt{1-x^2}} \] - The derivative of \( v(x) = \ln(x) \) is: \[ v'(x) = \frac{1}{x} \] Now, apply the product rule: \[ h'(x) = u'(x)v(x) + u(x)v'(x) \] \[ h'(x) = \left(-\frac{1}{\sqrt{1-x^2}}\right) \ln(x) + \arccos(x) \left(\frac{1}{x}\right) \] So, the derivative \( h'(x) \) is: \[ h'(x) = -\frac{\ln(x)}{\sqrt{1-x^2}} + \frac{\arccos(x)}{x} \] This expression represents the derivative of the given function \( h(x) = (\arccos(x)) \ln(x) \).
Transcribed Image Text:**Problem Statement:** Find the derivative of the function. Given: \[ h(x) = (\arccos(x)) \ln(x) \] **Solution:** To find the derivative \( h'(x) \), we will use the product rule, which states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product is given by: \[ (uv)' = u'v + uv' \] In this case, let: \[ u(x) = \arccos(x) \] \[ v(x) = \ln(x) \] First, find the derivatives of \( u(x) \) and \( v(x) \): - The derivative of \( u(x) = \arccos(x) \) is: \[ u'(x) = -\frac{1}{\sqrt{1-x^2}} \] - The derivative of \( v(x) = \ln(x) \) is: \[ v'(x) = \frac{1}{x} \] Now, apply the product rule: \[ h'(x) = u'(x)v(x) + u(x)v'(x) \] \[ h'(x) = \left(-\frac{1}{\sqrt{1-x^2}}\right) \ln(x) + \arccos(x) \left(\frac{1}{x}\right) \] So, the derivative \( h'(x) \) is: \[ h'(x) = -\frac{\ln(x)}{\sqrt{1-x^2}} + \frac{\arccos(x)}{x} \] This expression represents the derivative of the given function \( h(x) = (\arccos(x)) \ln(x) \).
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To find the derivative of the following function.

straight h open parentheses straight x close parentheses equals open parentheses arccos open parentheses straight x close parentheses close parentheses space ln open parentheses straight x close parentheses

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