Question

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(x^{\cos x}\right)$$

Short answer

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

Step-by-step solution

Take the natural logarithm of both sides

To apply logarithmic differentiation, we first take the natural logarithm of both sides of the function: $$\ln\left(y\right) = \ln\left(x^{\cos x}\right)$$ Where y is equal to the given function, i.e., $$y = x^{\cos x}$$

Apply the logarithmic property to simplify the equation

After taking the natural logarithm, we can use the logarithmic property $$\ln\left(a^b\right) = b\ln\left(a\right)$$ to simplify the right side of the equation: $$\ln\left(y\right) = \cos x \ln x$$

Differentiate both sides with respect to x

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

Apply the product rule to differentiate the right side

To differentiate the right side of the equation, we apply the product rule: $$\frac{d}{dx}\left(uv\right) = u\frac{dv}{dx} + v\frac{du}{dx}$$ where u is $$\cos x$$ and v is $$\ln x$$: $$\frac{1}{y}\frac{dy}{dx} = \cos x \frac{d}{dx}\left(\ln x\right) + \ln x \frac{d}{dx}\left(\cos x\right)$$

Differentiate the individual functions

Now, differentiate $$\ln x$$ and $$\cos x$$ with respect to x: $$\frac{d}{dx}\left(\ln x\right) = \frac{1}{x}$$ $$\frac{d}{dx}\left(\cos x\right) = -\sin x$$

Substitute the individual derivatives back into the equation

Replace the derivatives in the equation: $$\frac{1}{y}\frac{dy}{dx} = \cos x \left(\frac{1}{x}\right) - \ln x \left(\sin x\right)$$

Solve for the derivative dy/dx

Now, multiply both sides of the equation by y, which is equal to $$x^{\cos x}$$: $$\frac{dy}{dx} = \cos x\frac{1}{x}\left(x^{\cos x}\right) - \ln x \sin x\left(x^{\cos x}\right)$$

Simplify the result

Finally, simplify the expression by factoring out $$x^{\cos x}$$: $$\frac{dy}{dx} = x^{\cos x}\left(\cos x\frac{1}{x} - \ln x \sin x\right)$$ Now we have found the derivative of the given function: $$\frac{d}{d x}\left(x^{\cos x}\right) = x^{\cos x}\left(\cos x\frac{1}{x} - \ln x \sin x\right)$$

Key concepts

Calculus

Calculus is an area of mathematics that deals with changes. Two main branches define calculus: differential calculus and integral calculus. Differential calculus concerns the rates at which quantities change, while integral calculus focuses on accumulation of quantities, such as areas under a curve. The derivative, a fundamental tool in calculus, measures how a function changes as its input changes. This tool is essential for understanding motion, optimization problems, and rates of change in various fields of science and engineering.

In solving problems that involve complex functions, such as those with variable exponents, calculus offers various techniques, one of which is logarithmic differentiation. This method simplifies the process of differentiating functions that would be challenging to handle with basic rules of differentiation. As exemplified in the problem exercise, calculus concepts not only give us the tools to solve mathematical problems but also enhance our understanding of the world by modeling and solving real-world situations.

Exponential Derivatives

Exponential derivatives are derivatives of functions that have variable exponents, such as the function \(x^{\cos x}\) from the given exercise. Calculating these types of derivatives can be complex due to the variable exponent. To handle this complexity, we use the rules of logarithms and differentiation to make the process easier.

The reason we often resort to logarithmic differentiation for exponential functions is because logarithms convert multiplication into addition and powers into products, which are much simpler operations to differentiate. By taking the natural logarithm, we leverage these properties to transform a tricky exponential derivative into a more manageable form. This strategy is powerful especially when dealing with exponents that are themselves functions, a situation quite common in advanced calculus problems.

Product Rule

Understanding the Product Rule

The product rule is a basic rule in calculus used when differentiating products of two or more functions. Stated formally, if we have two differentiable functions \(u(x)\) and \(v(x)\), the derivative of their product is given by: \[\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}\].

In the context of the exercise, the product rule is applied after taking the natural logarithm of the function. When differentiating the product of \(\cos x\) and \(\ln x\), we used the product rule to find the derivative of each portion separately, then combined the results. This allows us to handle the differentiation of complex expressions by breaking them down into simpler parts, demonstrating the utility of the product rule in calculus.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is an essential concept not just in calculus, but in mathematics as a whole. It is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm has special properties that make it particularly useful when working with exponential functions.

One of the most important properties of natural logarithm is that the derivative of \(\ln x\) with respect to \(x\) is \(1/x\). This property simplifies our exercise problem, as taking the natural logarithm of the function \(x^{\cos x}\) allowed us to bring down the variable exponent and then differentiate more easily. This is a clear demonstration of how the natural logarithm functions as a bridge between exponential and polynomial type functions, providing a versatile tool for solving a variety of calculus problems.