Problem 52 - Calculus Early Transcendentals

Question

Compute the following derivatives using the method of your choice. $$\frac{d}{d x}\left(x^{\tan x}\right)$$

Step-by-Step Solution

Answer

Question: Compute the derivative of the function \(x^{\tan x}\) with respect to \(x\). Answer: \(\frac{d}{dx}(x^{\tan x}) = x^{\tan x} \cdot \left((\tan x)\cdot \frac{1}{x} + (\ln x)\cdot (\sec^2 x)\right)\)

Step 1: Take the natural logarithm of both sides

Since we are given the function \(x^{\tan x}\), we will take the natural logarithm of both sides which gives us: $$\ln(y) = \ln\left(x^{\tan x}\right)$$ where we let \(y = x^{\tan x}\).

Step 2: Differentiate both sides with respect to \(x\)

Now we will differentiate both sides of the equation with respect to \(x\): $$\frac{d}{dx}(\ln(y)) = \frac{d}{dx}\left(\ln\left(x^{\tan x}\right)\right)$$ Using the chain rule on the left side and the product rule on the right side, we get: $$\frac{1}{y}\cdot\frac{dy}{dx} = (\tan x)\cdot \frac{d}{dx}( \ln x) + (\ln x)\cdot \frac{d}{dx}( \tan x)$$ Now, we can use the known derivatives: \(\frac{d}{dx}(\ln x) = \frac{1}{x}\) and \(\frac{d}{dx}(\tan x) = \sec^2 x\). Substitute these into the equation: $$\frac{1}{y}\cdot\frac{dy}{dx} = (\tan x)\cdot \frac{1}{x} + (\ln x)\cdot (\sec^2 x)$$

Step 3: Solve for the derivative

Now, we will solve for the derivative \(\frac{dy}{dx}\) and remember that \(y = x^{\tan x}\): $$\frac{dy}{dx} = y \cdot \left((\tan x)\cdot \frac{1}{x} + (\ln x)\cdot (\sec^2 x)\right)$$ Substitute the value of \(y\) back into the equation: $$\frac{dy}{dx} = x^{\tan x} \cdot \left((\tan x)\cdot \frac{1}{x} + (\ln x)\cdot (\sec^2 x)\right)$$ This is the derivative of the given function: $$\boxed{\frac{d}{dx}(x^{\tan x}) = x^{\tan x} \cdot \left((\tan x)\cdot \frac{1}{x} + (\ln x)\cdot (\sec^2 x)\right)}$$

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithmic function with the base of the mathematical constant \( e \), approximately equal to 2.71828. It is a powerful tool for solving various exponential equations, especially in calculus.

In the context of derivatives, the natural logarithm can help simplify complex expressions. For example, taking the logarithm of both sides of an equation transforms multiplication into addition. This property is extremely useful when differentiating functions of the form \( f(x) = a^{g(x)} \). By using \( \ln(x) \), exponential functions can be handled algebraically much easier.

It is important to remember that the derivative of \( \ln(x) \) is straightforward: \( \frac{d}{dx}(\ln x) = \frac{1}{x} \). This allows us to play around with these functions and their transformations to find derivatives more efficiently.

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of a composite function. A composite function is when one function is applied inside of another, such as \( f(g(x)) \). The chain rule states that the derivative of \( f(g(x)) \) is the derivative of \( f \) evaluated at \( g(x) \) times the derivative of \( g \), or mathematically:

\[ \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \]

In our original problem, the chain rule is used to differentiate \( \ln(y) = \ln(x^{\tan x}) \). The left side becomes \( \frac{1}{y} \cdot \frac{dy}{dx} \), using the chain rule, considering \( y \) as a function dependent on \( x \). This approach makes differentiating composite and complicated functions much easier, providing a step-by-step route to simplify and compute their derivatives.

Product Rule

The product rule is essential for differentiating the product of two functions. Suppose you have two functions, \( u(x) \) and \( v(x) \), and you want to differentiate their product. The product rule formula is:

\[ \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \]

In the problem we tackled, the product rule came in handy when differentiating the right side of the equation: \( \ln\left(x^{\tan x}\right) \). Here, we encounter a product \( \tan x \cdot \ln x \). Applying the product rule, it is split into \( (\tan x)' \cdot \ln x + \tan x \cdot (\ln x)' \).

Using known derivatives like \( (\tan x)' = \sec^2 x \) and \( (\ln x)' = \frac{1}{x} \), we can plug these values in to find the complete product derivative. The product rule is indispensable because it ensures that we deal with interactions between functions accurately when they are being multiplied.

Trigonometric Derivatives

Trigonometric derivatives are derivatives of trigonometric functions like \( \sin x \), \( \cos x \), and \( \tan x \). Knowing these derivatives is vital as they frequently appear in calculus problems. Common derivatives are as follows:

• \( \frac{d}{dx}(\sin x) = \cos x \)
• \( \frac{d}{dx}(\cos x) = -\sin x \)
• \( \frac{d}{dx}(\tan x) = \sec^2 x \)

In the original problem, figuring out \( \frac{d}{dx}(\tan x) \) is necessary due to its appearance in the function \( x^{\tan x} \). We used this derivative when we applied the product rule to differentiate \( \ln(x)^{\tan x} \). The derivative \( \sec^2 x \) results from the fact that \( \sec x \) (or secant) is the reciprocal of the cosine function.

Mastery of these trigonometric derivatives allows for seamless manipulation and understanding of calculus problems involving trigonometric functions.