Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Problem Statement:** Given that: \( f(x) = x^7 h(x) \) \( h(-1) = 5 \) \( h'(-1) = 8 \) Calculate \( f'(-1) \). **Solution:** To solve this, we use the product rule for derivatives. The product rule states that if \( f(x) = u(x)v(x) \), then the derivative \( f'(x) \) is: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] In this case, let \( u(x) = x^7 \) and \( v(x) = h(x) \). 1. Calculate \( u'(x) \): \[ u(x) = x^7 \] \[ u'(x) = 7x^6 \] 2. Evaluate \( u'(-1) \): \[ u'(-1) = 7(-1)^6 = 7 \] 3. Plug in the values into the product rule: \[ f'(-1) = u'(-1)v(-1) + u(-1)v'(-1) \] Substituting the given values: \[ f'(-1) = 7 \cdot 5 + (-1)^7 \cdot 8 \] \[ f'(-1) = 35 - 8 \] \[ f'(-1) = 27 \] Thus, \( f'(-1) = 27 \).
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Transcribed Image Text:**Problem Statement:** Given that: \( f(x) = x^7 h(x) \) \( h(-1) = 5 \) \( h'(-1) = 8 \) Calculate \( f'(-1) \). **Solution:** To solve this, we use the product rule for derivatives. The product rule states that if \( f(x) = u(x)v(x) \), then the derivative \( f'(x) \) is: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] In this case, let \( u(x) = x^7 \) and \( v(x) = h(x) \). 1. Calculate \( u'(x) \): \[ u(x) = x^7 \] \[ u'(x) = 7x^6 \] 2. Evaluate \( u'(-1) \): \[ u'(-1) = 7(-1)^6 = 7 \] 3. Plug in the values into the product rule: \[ f'(-1) = u'(-1)v(-1) + u(-1)v'(-1) \] Substituting the given values: \[ f'(-1) = 7 \cdot 5 + (-1)^7 \cdot 8 \] \[ f'(-1) = 35 - 8 \] \[ f'(-1) = 27 \] Thus, \( f'(-1) = 27 \).
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