Question

Logarithmic differentiation Use logarithmic differentiation to evaluate \(f^{\prime}(x)\). $$f(x)=x^{\left(x^{10}\right)}$$

Short answer

Question: Use logarithmic differentiation to find the derivative of the function \(f(x) = x^{\left(x^{10}\right)}\). Answer: The derivative of the given function, \(f'(x)\), is given by: $$f'(x)=x^{\left(x^{10}\right)}\cdot \left(10x^9\ln(x)+x^{10}\cdot \frac{1}{x}\right)$$

Step-by-step solution

Take the natural logarithm of both sides of the equation

To apply logarithmic differentiation, we'll start by taking the natural logarithm on both sides of the function: $$\ln(f(x))=\ln\left(x^{\left(x^{10}\right)}\right)$$

Apply logarithm properties

We can use the logarithmic property of exponents to simplify the right side of the equation: $$\ln(f(x))=x^{10}\ln(x)$$

Differentiate both sides of the equation with respect to x

Now we can differentiate both sides with respect to x: $$\frac{f'(x)}{f(x)}=10x^9\ln(x)+x^{10}\cdot \frac{1}{x}$$

Solve for \(f'(x)\)

To find the derivative, \(f'(x)\), we need to multiply both sides by \(f(x)\): $$f'(x)=f(x)\cdot \left(10x^9\ln(x)+x^{10}\cdot \frac{1}{x}\right)$$

Substitute \(f(x)\) back into the expression

Recall that \(f(x) = x^{\left(x^{10}\right)}\). Substitute \(f(x)\) into the expression: $$f'(x)=x^{\left(x^{10}\right)}\cdot \left(10x^9\ln(x)+x^{10}\cdot \frac{1}{x}\right)$$ Now, we have found the derivative of the given function using logarithmic differentiation.

Key concepts

Natural Logarithm

Understanding the natural logarithm is essential in calculus, especially when dealing with complex functions that require differentiation. The natural logarithm, denoted as \(\ln(x)\), is the logarithm to the base \(e\), where \(e\) is the irrational constant approximately equal to 2.71828. This special logarithm is important because it has properties that make it easier to work with, particularly when evaluating the rates of growth or decay.

In calculus, \(\ln\) is often used to simplify multiplication into addition or exponentiation into multiplication, making complex functions easier to differentiate. It's also the inverse function of \(e^x\), which means for any \(x\), \(e^{\ln(x)} = x\). The natural logarithm's derivative is also significant; the derivative of \(\ln(x)\) with respect to \(x\) is \(1/x\), which plays a crucial role in solving various calculus problems.

Derivatives

A derivative represents the rate at which a function is changing at any given point, and it's a fundamental concept in calculus. Intuitively, the derivative of a function at a point is the slope of the tangent line to the function's graph at that point. Derivatives allow us to understand motion, growth, and change across a multitude of applications.

When calculating a derivative, you're essentially finding a new function that gives the slope of the original function at any given \(x\). Notation-wise, if the original function is \(f(x)\), then its derivative is represented by \(f'(x)\) or \(\frac{df}{dx}\). The process of finding a derivative is called differentiation, and there are various rules and methods, including the power rule, product rule, quotient rule, and chain rule, to simplify and facilitate this process. Logarithmic differentiation, as in our example problem, is a technique to handle functions that are difficult to differentiate directly.

Logarithmic Properties

Using logarithmic properties can greatly simplify the process of differentiation, especially when dealing with products, quotients, or powers that would otherwise be cumbersome. There are several key properties of logarithms:

• Product property: \(\ln(ab) = \ln(a) + \ln(b)\), which turns multiplication into addition.
• Quotient property: \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\), which transforms division into subtraction.
• Power property: \(\ln(a^b) = b\ln(a)\), allowing us to turn exponentiation into multiplication.
• Change of base formula: \(\log_b a = \frac{\ln(a)}{\ln(b)}\), which is useful for converting between logarithm bases.

By applying these properties, we can rewrite complex logarithmic expressions in a form that is more straightforward to differentiate, as demonstrated in the exercise with logarithmic differentiation.