Question

Evaluate the following derivatives. $$\frac{d}{d x}\left(\left(\frac{1}{x}\right)^{x}\right)$$

Short answer

Question: Find the derivative of the function \(f(x)=\left(\frac{1}{x}\right)^x\) with respect to x. Answer: The derivative of the function is \(\frac{d}{dx}(\left(\frac{1}{x}\right)^x) = x^{-x}(-1-\ln{x})\).

Step-by-step solution

Rewrite the function

We can rewrite the given function as: $$y = x^{-x}$$ This will make it easier to apply logarithmic differentiation.

Apply natural logarithm to both sides

Now, apply the natural logarithm to both sides of the equation: $$\ln{y} = \ln{(x^{-x})}$$

Simplify the right-hand side

Using the properties of logarithm, we can bring the exponent down: $$\ln{y} = -x\ln{x}$$

Differentiate implicitly with respect to x

Now, differentiate both sides of the equation with respect to x: $$\frac{1}{y}\frac{dy}{dx}=-x\frac{d}{dx}\ln{x}-\ln{x}\frac{d}{dx}x$$

Compute the derivatives on the right-hand side

The derivative of \(\ln{x}\) can be found by using the chain rule: $$\frac{d}{dx}(\ln{x}) = \frac{1}{x}$$ And the derivative of \(x\) is 1: $$\frac{d}{dx}(x) = 1$$ Substituting these back into our previous expression: $$\frac{1}{y}\frac{dy}{dx}=-x\frac{1}{x}-\ln{x}(1)$$

Simplify further

The x terms cancel out on the right-hand side: $$\frac{1}{y}\frac{dy}{dx}=-1-\ln{x}$$

Isolate the derivative

Now, multiply both sides by \(y\) to isolate the derivative: $$\frac{dy}{dx}=y(-1-\ln{x})$$

Substitute the original function for y

Finally, substitute the original function (from Step 1) for \(y\): $$\frac{dy}{dx}=x^{-x}(-1-\ln{x})$$ Thus, the derivative of the given function is: $$\frac{d}{d x}\left(\left(\frac{1}{x}\right)^{x}\right) = x^{-x}(-1-\ln{x})$$

Key concepts

Calculus

Calculus is the mathematical study of continuous change. It is a branch of mathematics that includes two related ideas: differentiation and integration. Differentiation focuses on rates of change and slopes of curves, which helps us understand motion, growth, and decay. Integration, on the other hand, deals with finding the total size or value, such as areas under curves and accumulated quantities. Calculus is not just a subject; it's a tool that enables us to solve complex real-world problems by breaking them down into simpler parts.

The field of calculus has been developed over centuries and is fundamental in various domains including physics, engineering, economics, statistics and more. Whether we are explaining the trajectory of a planet or optimizing a business model, calculus offers the precision and tools necessary to articulate and solve such problems.

Derivatives

In calculus, derivatives represent the rate at which one quantity changes with respect to another. If we have a function, say, involving time and distance, the derivative of this function would tell us the speed – how quickly the distance changes as time progresses.

The process of finding a derivative is called differentiation. The rules of differentiation are grounded in limits and can be used to find the slope of a tangent line to a curve at a point. This is essential for understanding the behavior of functions. Derivatives are not only about velocities and inclines but also about finding the rates of change in a wide array of physical, biological, and economic contexts.

Chain Rule

The chain rule is a fundamental technique in calculus for taking the derivative of the composition of two or more functions. Whenever we have a function applied inside another function, the chain rule guides us on how to differentiate this compound setup effectively.

To use the chain rule, you identify the outer function and the inner function within a composite function. You then differentiate the outer function with respect to the inner function and multiply that by the derivative of the inner function with respect to the independent variable. Mathematically, if we have a function \(h(x) = f(g(x))\), the chain rule tells us that \(h'(x) = f'(g(x)) \cdot g'(x)\).

Implicit Differentiation

While many functions are written in an explicit format, like \(y = f(x)\), some are not easily solvable for \(y\) and are presented in an implicit form, where \(y\) and \(x\) are mingled together. Implicit differentiation helps us find the derivative without explicitly solving for one variable in terms of the other.

This technique is especially relevant when dealing with more complex equations which involve products and powers of \(x\) and \(y\) intermixed. To perform implicit differentiation, we differentiate both sides of the equation with respect to \(x\), treating \(y\) as an implicit function of \(x\), and then solve for \(dy/dx\). This allows us to find the rate at which \(y\) changes with respect to \(x\) without having to isolate \(y\) first.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm is the inverse operation of exponentiation with base \(e\), meaning if \(y = \ln(x)\), then \(e^y = x\).

The natural logarithm is particularly important in calculus because of its unique properties, such as the derivative of \(\ln(x)\) being \(1/x\) and the integration of \(1/x\) yielding \(\ln|x|\) plus a constant. It plays a critical role in solving problems that involve growth and decay, compounding interest, and in many areas of science and mathematics. Understanding the natural logarithm is key when dealing with exponential functions and log differentiation.