Question

Evaluate the following derivatives. $$\frac{d}{d x}(\sin (\ln x))$$

Short answer

Answer: The derivative of the function \(f(x) = \sin (\ln x)\) with respect to \(x\) is \(\frac{d}{dx}(\sin (\ln x)) = \frac{\cos (\ln x)}{x}\).

Step-by-step solution

Identify the inner and outer functions

In order to apply the Chain Rule, we first need to identify the inner function \(g(x)\) and the outer function \(f(u)\). In this case: - Inner function: \(g(x) = \ln x\) - Outer function: \(f(u) = \sin u\), where \(u = g(x) = \ln x\)

Find the derivative of the inner function

Next, we'll compute the derivative of the inner function with respect to \(x\). The derivative of the natural logarithm function is: $$\frac{d}{dx}(\ln x) = \frac{1}{x}$$ So, \(g'(x) = \frac{1}{x}\).

Find the derivative of the outer function with respect to \(u\)

Now we'll find the derivative of the outer function \(f(u)\) with respect to \(u\). Since \(f(u) = \sin u\), the derivative is: $$\frac{d}{du}(\sin u) = \cos u$$ However, recall that \(u = \ln x\). So, the derivative can also be written as \(f'(u) = \cos (\ln x)\).

Apply the Chain Rule

Finally, we can apply the Chain Rule to find the derivative \(\frac{d}{dx}(\sin(\ln(x)))\) by multiplying the derivatives from steps 2 and 3: $$\frac{d}{dx}[\sin(\ln x)] = f'(g(x))g'(x) = \cos (\ln x) \cdot \frac{1}{x}$$ Therefore, the derivative of the function is: $$\frac{d}{dx}(\sin (\ln x)) = \frac{\cos (\ln x)}{x}$$

Key concepts

Derivative of Natural Logarithm

Understanding the derivative of the natural logarithm function is crucial for higher-level calculus. The natural logarithm, denoted as \(\ln x\), is the inverse function of the exponential \(e^x\). The derivative of \(\ln x\) is given by the formula:
\[\frac{d}{dx}(\ln x) = \frac{1}{x}\]
This signifies that the rate at which \(\ln x\) changes with respect to \(x\) is proportional to the reciprocal of \(x\). This fundamental concept forms a cornerstone for tackling more complex differentiation problems, especially when combined with other functions through composition, as seen in the Chain Rule application.

Derivative of Sine Function

The sine function, \(\sin x\), is a periodic function commonly used to model oscillations and waves. In calculus, the derivative of the sine function is essential for analyzing the rate of change in oscillatory systems. The derivative is given by:
\[\frac{d}{dx}(\sin x) = \cos x\]
This means that the slope of the sine function at any point is equal to the value of the cosine function at that same point. As we venture into the realm of trigonometric derivatives, recognizing this relationship between sine and cosine functions is invaluable for mastering differentiation techniques.

Applying the Chain Rule in Calculus

When dealing with the differentiation of composite functions, the Chain Rule is the tool of choice. The basic strategy of the Chain Rule is to break down a complex function into its component functions and differentiate each part systematically.
When given a composite function \(f(g(x))\), where \(g(x)\) is the inner function and \(f(u)\) is the outer function, the Chain Rule states that:
\[\frac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)\]
Here, \(f'(g(x))\) is the derivative of the outer function evaluated at \(g(x)\), and \(g'(x)\) is the derivative of the inner function with respect to \(x\). When applied correctly, the Chain Rule allows us to navigate the differentiation of even the most intricate functions.

Composite Functions Differentiation

Differentiation of composite functions is an extension of the basic principles of differentiation to functions that incorporate other functions within them. It is common to think of these as functions inside of functions.
For instance, if you have a function \(h(x) = f(g(x))\), \(h(x)\) is a composite function, where \(g(x)\) is nested within \(f\). The process involves taking the derivative of the outer function with respect to the inner function and then multiplying it by the derivative of the inner function with respect to \(x\). This layered approach simplifies complex expressions, making it possible to find the rate of change for an array of intricate relationships between variables.