Problem 40 - Single Variable Calculus

Question

Use logarithmic differentiation to find the derivative of the function. $$ y=\sin x^{\tan x} $$

Step-by-Step Solution

Answer

The derivative of the function \(y=\sin(x^{\tan x})\) is: \[y' = \sin(x^{\tan x}) \cdot \left[ x^{\tan x} \cdot ((\sec^2 x) \cdot \ln x + \frac{\tan x}{x} )\right]\]

Step 1: Take the natural logarithm of both sides

Since we want to use logarithmic differentiation, let's first take the natural logarithm of both sides of the equation. This gives: \[\ln(y) = \ln(\sin(x^{\tan x}))\]

Step 2: Differentiate both sides with respect to x

Now, we will differentiate both sides of the equation with respect to x. On the left side, use the chain rule (derivative of ln(y) times derivative of y with respect to x). On the right side, we will first differentiate the outer function \(\ln(\cdot)\) and use the chain rule to multiply it by the derivative of the inner function \(\sin(x^{\tan x})\). The inner function can be differentiated using the chain rule as well. \[\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\ln(\sin(x^{\tan x}))\] Differentiating the right side, we get: \[\frac{1}{y}\frac{dy}{dx} = \frac{\cos(x^{\tan x})\cdot d(x^{\tan x})}{\sin(x^{\tan x})}\] Now let's differentiate \(x^{\tan x}\) using logarithmic differentiation. First, take the natural logarithm: \[\ln(x^{\tan x}) = \tan x \cdot \ln x\] And differentiate both sides with respect to x: \[\frac{1}{x^{\tan x}} \cdot d(x^{\tan x}) = (\sec^2 x)\cdot \ln x + \frac{\tan x}{x}\] Now, substitute this back into our main equation to find the derivative of the function: \[\frac{1}{y}\frac{dy}{dx} = \frac{\cos(x^{\tan x})}{\sin(x^{\tan x})} \cdot \left[ x^{\tan x} \cdot ((\sec^2 x) \cdot \ln x + \frac{\tan x}{x}) \right]\]

Step 3: Solve for y'

Now we need to solve for \(\frac{dy}{dx}\) by multiplying both sides by y: \[\frac{dy}{dx} = \sin(x^{\tan x}) \cdot \left[x^{\tan x} \cdot ((\sec^2 x) \cdot \ln x + \frac{\tan x}{x}) \right]\] So, the derivative of the function \(y=\sin(x^{\tan x})\) is: \[\boxed{y' = \sin(x^{\tan x}) \cdot \left[ x^{\tan x} \cdot ((\sec^2 x) \cdot \ln x + \frac{\tan x}{x} )\right]}\]

Key Concepts

Chain Rule

Understanding the chain rule is essential for differentiating composite functions. In calculus, the chain rule is a formula that computes the derivative of a function composed of two or more functions. Imagine you have a function that is the result of another function, like a nesting doll. To find the rate of change of the entire setup, you apply the chain rule.

The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function itself. In the context of logarithmic differentiation, this is particularly useful when working with functions where one function is nested inside another, such as \( \sin(x^{\tan x}) \).

To apply the chain rule, identify the 'outer' function and the 'inner' function. In our example, the outer function is the logarithm function \( \ln \), and the inner function is the sine function with its exponent \( \sin(x^{\tan x}) \). Differentiate the outer function, apply the result to the inner function, and then multiply by the derivative of the inner function. This process can be repeated as many times as necessary for functions with multiple layers.

Natural Logarithm

The natural logarithm is a fundamental concept in mathematics, particularly when dealing with exponential and logarithmic functions. It is denoted as \( \ln \), and it's the inverse function of the exponential function \( e^x \), where \( e \) is Euler's number, approximately 2.71828. The natural logarithm transforms multiplicative processes into additive ones, which simplifies the differentiation and integration of exponential functions.

When applying logarithmic differentiation, taking the natural logarithm of both sides of an equation helps reduce complicated products, quotients, and powers into sums, differences, and products which can be more easily differentiated with respect to \( x \). For instance, in our problem, we used \( \ln(y) = \ln(\sin(x^{\tan x})) \) to set the stage for applying the rules of differentiation. It allows us to handle the exponent \( \tan x \) in a more manageable way by transforming it into a product after logarithmic differentiation.

Implicit Differentiation

Implicit differentiation comes into play when dealing with equations in which the dependent variable cannot be easily solved for and expressed explicitly as a function of the independent variable. In such cases, rather than explicitly solving for one variable in terms of the other, we differentiate both sides of the equation with respect to the independent variable, treating the dependent variable as an implicit function of the independent variable.

In our exercise, we used implicit differentiation when we differentiated both sides of the equation with respect to \( x \). We did not solve for \( y \) directly but instead found \( \frac{dy}{dx} \) as it relates to \( y \) itself. By treating \( y \) as a function of \( x \) (even though it's not explicitly solved), we were able to differentiate it appropriately and then solve for \( y' \), which is the derivative of \( y \). Implicit differentiation is particularly useful in logarithmic differentiation because it allows us to manage derivatives of functions that involve other functions of \( y \) itself.