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What is the value of the derivative of f(x) = ln(x^2 + 1) at x = 1?
What is the value of the derivative of f(x) = ln(x^2 + 1) at x = 1?
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What is the value of the derivative of f(x) = ln(x^2 + 1) at x = 1?
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f'(x) = (2x)/(x^2 + 1). At x = 1, f'(1) = 2/(1 + 1) = 1.
Questions & Step-by-step Solutions
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Q: What is the value of the derivative of f(x) = ln(x^2 + 1) at x = 1?
Solution:
f'(x) = (2x)/(x^2 + 1). At x = 1, f'(1) = 2/(1 + 1) = 1.
Steps: 8
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Step 1: Identify the function we need to differentiate. The function is f(x) = ln(x^2 + 1).
Step 2: Use the chain rule to find the derivative of f(x). The derivative of ln(u) is (1/u) * (du/dx), where u = x^2 + 1.
Step 3: Calculate du/dx. Since u = x^2 + 1, the derivative du/dx = 2x.
Step 4: Substitute u and du/dx into the derivative formula. We get f'(x) = (1/(x^2 + 1)) * (2x).
Step 5: Simplify the expression for the derivative. This gives us f'(x) = (2x)/(x^2 + 1).
Step 6: Now, we need to find the value of the derivative at x = 1. Substitute x = 1 into f'(x).
Step 7: Calculate f'(1) = (2*1)/(1^2 + 1) = 2/(1 + 1) = 2/2 = 1.
Step 8: The value of the derivative of f(x) at x = 1 is 1.
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