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Magazine Chapter 3: Problem 24
Find \(d y / d x\) $$ y=\sin ^{2}(\ln x) $$
Short Answer
Expert verified
\( \frac{d y}{d x} = \frac{2 \sin(\ln x) \cos(\ln x)}{x} \)
Step by step solution
01
Identify the Composition
The given function is \( y = \sin^2(\ln x) \). It's a composition involving a power, a trigonometric function, and a logarithmic function. Specifically, it is the square of a sine function which itself contains a logarithmic function.
02
Apply Chain Rule
Use the chain rule to differentiate the composite function. Start by letting \( u = \ln x \), then \( v = \sin u = \sin(\ln x) \). Therefore, \( y = v^2 = (\sin(\ln x))^2 \). We need to find \( d y / d x \) by finding the derivatives of \( u \) and \( v \) with respect to \( x \).
03
Differentiate the Outer Function
Since \( y = v^2 \), use the chain rule: \( \frac{d y}{d v} = 2v \).
04
Differentiate the Inner Function
Differentiate \( v = \sin u \) with respect to \( u \): \( \frac{d v}{d u} = \cos u = \cos(\ln x) \).
05
Differentiate the Innermost Function
Differentiate \( u = \ln x \) with respect to \( x \): \( \frac{d u}{d x} = \frac{1}{x} \).
06
Combine Derivatives Using Chain Rule
By the chain rule, \( \frac{d y}{d x} = \frac{d y}{d v} \cdot \frac{d v}{d u} \cdot \frac{d u}{d x} = 2v \cdot \cos(\ln x) \cdot \frac{1}{x} \). Now substitute \( v = \sin(\ln x) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule in Derivative Calculus
The chain rule is a fundamental technique in calculus used to differentiate composite functions. When you encounter a function inside another function, the chain rule helps you find the derivative. In our example, the composite function is composed of the natural logarithm, the sine function, and a squared operation. This requires the use of the chain rule multiple times.
- Start by recognizing inner and outer functions. For example, if you have a function like \( y = f(g(x)) \), you need to differentiate \( f \) and \( g \) separately.
- For the outer function \( f \), once you've differentiated it with respect to its inner function, you'll multiply by the derivative of the inner function \( g \).
- First, differentiate \( y = v^2 \) concerning \( v \).
- Then differentiate \( v = \sin(u) \) concerning \( u \).
- Finally, differentiate \( u = \ln(x) \) concerning \( x \).
Composite Functions Made Simple
A composite function is simply a function within another function. It's like a nested layer of operations where one function operates on the results of another. This Nested structure is evident in our example function: \( y = \sin^2(\ln x) \).
- The innermost function here is \( \ln x \), which transforms \( x \) to its logarithm.
- Then, \( \sin(\ln x) \) modifies this result by applying the sine function.
- Finally, the squared operation squares the outcome from the sine function.
Logarithmic Differentiation
Logarithmic differentiation is a clever technique used when differentiating functions involving logarithms. It's especially useful when the function includes products or powers. In the given problem, because we have a logarithmic function contained within a trigonometric and a power function, it's vital to understand how to differentiate \( \ln x \).
- The natural logarithm function, \( \ln x \), plays a significant role in simplifying the differentiation process due to its straightforward derivative.
- The derivative of \( \ln x \) is simply \( \frac{1}{x} \). This is simple but crucial when linked with other functions.
