91Ó°ÊÓ

Derivatives of power functions are a foundational concept in calculus, particularly when dealing with functions of the form \( f(x) = x^n \) where \( n \) is a constant. The basic rule states that the derivative of \( x^n \) is \( nx^{n-1} \). But what happens when both the base and the exponent are functions of \( x \) instead of simple constants? This is where logarithmic differentiation becomes useful.

Logarithmic differentiation is a method used to differentiate functions that are not easily handled by standard rules of differentiation. It simplifies the differentiation process by taking the natural logarithm of both sides of the equation and then applying differentiation rules. In the given exercise, \( f(x) = (x^2+1)^x \) is not a basic power function since the exponent \( x \) is also a variable.

Using logarithmic differentiation, we transform the power function into a format that allows us to apply the derivatives of power functions and other rules like the product and chain rules. This method reveals the underlying structure of the function, making it easier to manipulate and differentiate.

The chain rule is a pivotal tool in calculus employed to compute the derivative of a composite function. When we have two or more functions combined, say \( u(v(x)) \), the derivative of this composition is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. It essentially chains the derivatives of the individual functions together.

In other words, if we have \( f(x) = u(v(x)) \), then the derivative \( f'(x) \) is \( u'(v(x)) \cdot v'(x) \). In the solution provided for the function \( f(x) = (x^2+1)^x \), the chain rule applies after logarithmic differentiation has reduced the function to \( x\ln(x^2+1) \). Differentiating \( \ln(x^2+1) \) requires us to differentiate its 'inner function' \( x^2+1 \) which showcases the chain rule’s ability to handle the complexity of nested functions.

The product rule is essential when we need to differentiate expressions where two or more functions are multiplied together. The rule states that if we have two functions, \( u(x) \) and \( v(x) \) that are multiplied together to form \( f(x) = u(x) \cdot v(x) \), then the derivative \( f'(x) \) is \( u'(x)v(x) + u(x)v'(x) \). This means we take the derivative of the first function and multiply it by the second function as is, then add the product of the first function as is and the derivative of the second function.

When applying the product rule in logarithmic differentiation, as seen in the exercise \( x\ln(x^2+1) \), the derivative \( \frac{f'(x)}{f(x)} = \ln(x^2+1) + \frac{2x^2}{x^2+1}\cdot x \) emerges. We distinguish the two functions \( u(x) = \ln(x^2+1) \) and \( v(x) = x \) and apply the product rule to find \( f'(x) \) in relation to \( f(x) \). The final step is multiplying both sides by \( f(x) \) to isolate the derivative \( f'(x) \) on one side, a maneuver illustrating the product rule’s versatility in complex differentiation scenarios.