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Magazine Chapter 3: Problem 347
For the following exercises, use logarithmic differentiation to find \(\frac{d y}{d x}\) $$y=(\sin 2 x)^{4 x}$$
Short Answer
Expert verified
\(\frac{dy}{dx} = (\sin 2x)^{4x}(4 \ln(\sin 2x) + 8x \cot 2x)\)
Step by step solution
01
Recognize the Function Form
The function given is \(y = (\sin 2x)^{4x}\). This is a power function where both the base and the exponent are functions of \(x\).
02
Apply Logarithmic Differentiation
Take the natural logarithm of both sides to simplify the differentiation process:\[\ln y = \ln((\sin 2x)^{4x})\]
03
Simplify Using Logarithm Properties
Utilize the power rule of logarithms to bring the exponent down:\[\ln y = 4x \ln(\sin 2x)\]
04
Differentiate Both Sides
Differentiate both sides of the equation with respect to \(x\). Use implicit differentiation for the left side and the product rule for the right side:\[\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[4x \ln(\sin 2x)]\]For the right side, apply the product rule:\[\frac{d}{dx}[4x \ln(\sin 2x)] = 4 \ln(\sin 2x) + 4x \cdot \frac{d}{dx}[\ln(\sin 2x)]\]
05
Differentiate \(\ln(\sin 2x)\)
Use the chain rule to differentiate \(\ln(\sin 2x)\):\[\frac{d}{dx}[\ln(\sin 2x)] = \frac{1}{\sin 2x} \cdot \cos 2x \cdot 2 = \frac{2\cos 2x}{\sin 2x} = 2 \cot 2x\]
06
Substitute Back into the Equation
Substitute the differentiated terms back into the expression for \( \frac{dy}{dx} \):\[\frac{1}{y} \frac{dy}{dx} = 4 \ln(\sin 2x) + 8x \cot 2x\]
07
Solve for \(\frac{dy}{dx}\)
Multiply both sides by \(y\) to find \(\frac{dy}{dx}\):\[\frac{dy}{dx} = y(4 \ln(\sin 2x) + 8x \cot 2x)\]Recall that \(y = (\sin 2x)^{4x}\), so:\[\frac{dy}{dx} = (\sin 2x)^{4x}(4 \ln(\sin 2x) + 8x \cot 2x)\]
08
Finalize the Solution
The derivative \(\frac{dy}{dx}\) is given by:\[\frac{dy}{dx} = 4x (\sin 2x)^{4x - 1} (2 \cos 2x) + (\sin 2x)^{4x} \cdot \ln(\sin 2x)\] \left[\text{Simplification and Correct Expression Form}\right]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative Calculus
Derivative calculus is a cornerstone of mathematical analysis, helping understand how functions change. It involves finding the derivative, which is the rate at which a function is changing at a given point. In our exercise, we are specifically interested in the derivative of a function that is a mix of both power and transcendental components.
- The original function is expressed as \[y = (\sin 2x)^{4x}\]
- To derive this, we use logarithmic differentiation, a technique handy when dealing with complicated products, quotients, or when variables appear in exponents.
Implicit Differentiation
Implicit differentiation is a technique that allows us to find the derivative of functions that aren't in the form \(y=f(x)\). Instead of isolating \(y\) on one side, we differentiate both sides of the equation concerning \(x\) and solve for \(\frac{dy}{dx}\) afterward.
- In step 4 of our solution, after taking the natural logarithm of both sides, we retain \(\ln y\) on the left, prompting implicit differentiation.
- The goal is to express the derivative by taking the differential of both sides: \[\frac{1}{y} \frac{dy}{dx} = \text{differentiated right side}\]
Chain Rule
The chain rule is a fundamental tool for differentiating composite functions鈥攆unctions within other functions. It helps determine how a change in a nested function affects the outer function.
- In our problem, see its application in the differentiation of \( \ln(\sin 2x) \).
- The chain rule formula is: \[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]
Product Rule
When a function is expressed as a product of two functions, the product rule comes into play. This rule helps us differentiate the product of two functions efficiently.
- To use the product rule, take two functions, \(u(x)\) and \(v(x)\); then the derivative of their product is given by \[\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)\]
- In our solution, we apply the product rule to \[4x \ln(\sin 2x)\] from the simplified logarithm, giving \[4 \ln(\sin 2x) + 4x \cdot \frac{2 \cos 2x}{\sin 2x}\]
