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Magazine Chapter 4: Problem 83
$$ \text { Differentiate. } $$ $$ y=(\sin x) \ln (\tan x) $$
Short Answer
Expert verified
The derivative is \[ y' = (\text{cos} \, x)(\ln(\tan x)) + \sec x \].
Step by step solution
01
Identify the function components
The function given is a product of two functions: \[ y = (\text{sin} \, x) \ln(\tan x) \]. Here, let鈥檚 identify the individual components. Let \( u = \text{sin} \, x \) and \( v = \ln(\tan x) \). The task is to differentiate the product \( y = u \, v \).
02
Apply the product rule
To differentiate a product of two functions, use the product rule: \( y' = u'v + uv' \). This requires the differentiation of both \( u \) and \( v \).
03
Differentiate \( u = \text{sin} \, x \)
Find the derivative of \( u \): \( u' = \frac{d}{dx}(\text{sin} \, x) = \text{cos} \, x \).
04
Differentiate \( v = \ln(\tan x) \)
Find the derivative of \( v \): \( v' = \frac{d}{dx}(\ln(\tan x)) \). Use the chain rule: \[ v' = \frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\sec^2 x}{\tan x} \].
05
Combine using the product rule
Now combine the derivatives using the product rule: \[ y' = u'v + uv' \]. Substituting the derivatives, \[ y' = (\text{cos} \, x)(\ln(\tan x)) + (\text{sin} \, x) \left( \frac{\sec^2 x}{\tan x} \right) \].
06
Simplify the expression
Simplify the expression: \[ y' = (\text{cos} \, x)(\ln(\tan x)) + (\text{sin} \, x) \left( \frac{1}{\cos^2 x \tan x} \right) = (\text{cos} \, x)(\ln(\tan x)) + \frac{\text{sin} \, x}{\sin x \cos x} = (\text{cos} \, x)(\ln(\tan x)) + \frac{1}{\cos x} \]. After further simplification, \[ y' = (\text{cos} \, x)(\ln(\tan x)) + \sec x \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
The product rule is a crucial differentiation technique used when you need to find the derivative of a product of two functions. Consider two functions, \( u(x) \) and \( v(x) \). According to the product rule, the derivative of their product is given by:
\[ (uv)' = u'v + uv' \]
Here's a step-by-step breakdown of how to apply this rule:
In our exercise, the two functions were \( u = \sin x \) and \( v = \ln(\tan x) \). We individually found their derivatives and then applied the product rule to get the final result.
\[ (uv)' = u'v + uv' \]
Here's a step-by-step breakdown of how to apply this rule:
- First, identify the two functions involved in the product.
- Next, find the derivatives of each of these functions individually.
- Finally, combine these derivatives using the product rule formula.
In our exercise, the two functions were \( u = \sin x \) and \( v = \ln(\tan x) \). We individually found their derivatives and then applied the product rule to get the final result.
Chain Rule
The chain rule is an essential tool in differentiation, especially useful when dealing with the composition of functions. It helps find the derivative of a function that is nested within another function. The chain rule can be stated as:
\[ (f(g(x)))' = f'(g(x)) \, g'(x) \]
Let's break down how to apply the chain rule:
In the given exercise, we used the chain rule to differentiate \( v = \ln(\tan x) \). We treated \( \ln(x) \) as the outer function and \( \tan x \) as the inner function. We then differentiated the outer function and multiplied by the derivative of the inner function:
\[ v' = \frac{1}{\tan x} \cdot \sec^2 x \]
Applying the chain rule step-by-step ensures we capture all necessary parts of the differentiation process.
\[ (f(g(x)))' = f'(g(x)) \, g'(x) \]
Let's break down how to apply the chain rule:
- Identify the outer function and the inner function within the composition.
- Differentiate the outer function, keeping the inner function unchanged.
- Multiply by the derivative of the inner function.
In the given exercise, we used the chain rule to differentiate \( v = \ln(\tan x) \). We treated \( \ln(x) \) as the outer function and \( \tan x \) as the inner function. We then differentiated the outer function and multiplied by the derivative of the inner function:
\[ v' = \frac{1}{\tan x} \cdot \sec^2 x \]
Applying the chain rule step-by-step ensures we capture all necessary parts of the differentiation process.
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent are fundamental in calculus, particularly when differentiating complex expressions. Here are some important derivatives of basic trigonometric functions that are useful to remember:
In our exercise, we had to differentiate \( \sin x \) and \( \tan x \) within the given function. Knowing these derivatives and how to apply them simplifies the differentiation process significantly.
We used:
Understanding how to differentiate trigonometric functions ensures you can handle even more complex expressions involving multiple functions.
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( \cos x \) is \( -\sin x \).
- The derivative of \( \tan x \) is \( \sec^2 x \).
In our exercise, we had to differentiate \( \sin x \) and \( \tan x \) within the given function. Knowing these derivatives and how to apply them simplifies the differentiation process significantly.
We used:
- \( \sin x \rightarrow \cos x \)
- \( \tan x \rightarrow \sec^2 x \)
Understanding how to differentiate trigonometric functions ensures you can handle even more complex expressions involving multiple functions.
