Find f′(1) when f(x) = x^(1/x).

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First, recognize that the function f(x) = x^(1/x) can be rewritten using the exponential and logarithmic functions for easier differentiation. Express it as f(x) = e^(ln(x^(1/x))).

Simplify the expression using the property of logarithms: ln(x^(1/x)) = (1/x) * ln(x). Therefore, f(x) = e^((1/x) * ln(x)).

To differentiate f(x), use the chain rule. Let u(x) = (1/x) * ln(x), so f(x) = e^(u(x)). The derivative f'(x) = e^(u(x)) * u'(x).

Find u'(x) by differentiating u(x) = (1/x) * ln(x). Use the product rule: u'(x) = d/dx(1/x) * ln(x) + (1/x) * d/dx(ln(x)).

Calculate the derivatives: d/dx(1/x) = -1/x^2 and d/dx(ln(x)) = 1/x. Substitute these into the expression for u'(x) and simplify. Finally, substitute back into f'(x) = e^(u(x)) * u'(x) and evaluate at x = 1 to find f'(1).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. In this context, finding f′(1) involves calculating the derivative of f(x) at the specific point x = 1.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It states that if a function y = g(u) is composed with u = f(x), then the derivative dy/dx can be found by multiplying the derivative of g with respect to u by the derivative of f with respect to x. This rule is essential when dealing with functions like f(x) = x^(1/x), which can be expressed in a composite form.

Logarithmic Differentiation

Logarithmic differentiation is a method used to differentiate functions that are products or quotients of variables raised to variable powers. By taking the natural logarithm of both sides of the equation, we can simplify the differentiation process. This technique is particularly useful for functions like f(x) = x^(1/x), allowing us to handle the exponent more easily when finding the derivative.