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The chain rule is a pivotal concept in calculus used to find the derivative of composite functions. Imagine you have a function inside another function, much like a set of nested boxes. To differentiate this composite function, the chain rule helps by letting us systematically unpack these boxes.

• Consider a function that can be written in the form of \(h(x) = f(g(x))\). Here, \(g(x)\) is the inner function and \(f(u)\) (where \(u = g(x)\)) is the outer function.
• The chain rule formula is \(h'(x) = f'(g(x)) \cdot g'(x)\).
• This means you first differentiate the outer function, \(f(u)\), with respect to \(u\), and then multiply it by the derivative of the inner function \(g(x)\) with respect to \(x\).

In our exercise, we apply the chain rule to differentiate \(f(x) = (\ln x)^2\). Here, \(g(x) = \ln x\) is the inner, and \(f(u) = u^2\) where \(u = g(x)\), is the outer function. By applying the chain rule, we first find the derivative of the outer function, resulting in \(2u\), and then multiply it by the derivative of the inner function, \(\frac{1}{x}\). Therefore, the derivative is \(f'(x) = 2\ln(x) \cdot \frac{1}{x}\).

Natural logarithms are a special kind of logarithm where the base is Euler's number, approximately \(2.718\). They are denoted by \(\ln\) and are fundamentally significant in calculus and exponential growth models.

• The function \(\ln x\) maps positive real numbers \(x\) to real numbers, providing the power to which \(e\) must be raised to yield \(x\).
• A key property of natural logarithms is \(\ln(1) = 0\), because \(e^0 = 1\).
• The derivative of \(\ln x\) with respect to \(x\) is a crucial calculus rule: \(\frac{d}{dx} (\ln x) = \frac{1}{x}\).

In the given exercise, understanding this derivative is essential for differentiating \(\ln^2 x\), where recognizing \(\ln(1) = 0\) is vital for evaluating the derivative at \(x = 1\).

Derivative evaluation is the process of calculating the slope of the tangent line at a specific point on the graph of a function. This is crucial for establishing whether a function is differentiable at that point.

• A function is said to be differentiable at a point if its derivative exists at that point and provides a finite, definitive value.
• In our exercise, after obtaining the derivative \(f'(x) = \frac{2\ln(x)}{x}\), we substitute \(x = 1\) to evaluate it.
• Substituting \(x = 1\), we figure out \(f'(1) = \frac{2 \cdot \ln(1)}{1}\). Since \(\ln(1) = 0\), it follows that \(f'(1) = 0\).

This evaluation shows that the function \(f(x) = \ln^2 x\) is differentiable at \(x = 1\), as it results in a tangible slope value of 0, verifying the function's smoothness at this point.