Suppose that g is a function that is differentiable at x = 8 and that g(8) = -4 and g' (8) = 3. Find h'(8). %3D h(x) = (x2 + 1)g(x) %3D h' (8) =
Suppose that g is a function that is differentiable at x = 8 and that g(8) = -4 and g' (8) = 3. Find h'(8). %3D h(x) = (x2 + 1)g(x) %3D h' (8) =
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:**
Suppose that \( g \) is a function that is differentiable at \( x = 8 \) and that \( g(8) = -4 \) and \( g'(8) = 3 \). Find \( h'(8) \).
Given:
\[ h(x) = (x^2 + 1)g(x) \]
Find:
\[ h'(8) = \, ? \]
**Solution:**
To find \( h'(x) \), we use the product rule for differentiation. The product rule states that if you have two functions, \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) is given by:
\[ (uv)' = u'v + uv' \]
Here, let:
- \( u(x) = x^2 + 1 \)
- \( v(x) = g(x) \)
Thus, \( u'(x) = 2x \) and the derivative of \( v(x) \) is \( g'(x) \).
Applying the product rule to \( h(x) = (x^2 + 1)g(x) \), we get:
\[ h'(x) = (x^2 + 1)g'(x) + (2x)g(x) \]
Now, substitute \( x = 8 \):
\[ h'(8) = (8^2 + 1)g'(8) + (2 \times 8)g(8) \]
Calculate each part:
- \( 8^2 + 1 = 64 + 1 = 65 \)
- \( g'(8) = 3 \)
- \( 2 \times 8 = 16 \)
- \( g(8) = -4 \)
Substitute these into the equation:
\[ h'(8) = 65 \cdot 3 + 16 \cdot (-4) \]
\[ h'(8) = 195 - 64 \]
\[ h'(8) = 131 \]
So, the derivative \( h'(8) \) is 131.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F87946fbe-26a8-44e7-b719-4f36f43dcbac%2F9a5e9b2b-15af-401b-bd43-425c90969420%2Fv4ks29b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
Suppose that \( g \) is a function that is differentiable at \( x = 8 \) and that \( g(8) = -4 \) and \( g'(8) = 3 \). Find \( h'(8) \).
Given:
\[ h(x) = (x^2 + 1)g(x) \]
Find:
\[ h'(8) = \, ? \]
**Solution:**
To find \( h'(x) \), we use the product rule for differentiation. The product rule states that if you have two functions, \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) is given by:
\[ (uv)' = u'v + uv' \]
Here, let:
- \( u(x) = x^2 + 1 \)
- \( v(x) = g(x) \)
Thus, \( u'(x) = 2x \) and the derivative of \( v(x) \) is \( g'(x) \).
Applying the product rule to \( h(x) = (x^2 + 1)g(x) \), we get:
\[ h'(x) = (x^2 + 1)g'(x) + (2x)g(x) \]
Now, substitute \( x = 8 \):
\[ h'(8) = (8^2 + 1)g'(8) + (2 \times 8)g(8) \]
Calculate each part:
- \( 8^2 + 1 = 64 + 1 = 65 \)
- \( g'(8) = 3 \)
- \( 2 \times 8 = 16 \)
- \( g(8) = -4 \)
Substitute these into the equation:
\[ h'(8) = 65 \cdot 3 + 16 \cdot (-4) \]
\[ h'(8) = 195 - 64 \]
\[ h'(8) = 131 \]
So, the derivative \( h'(8) \) is 131.
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