https://file.aiquickdraw.com/mathbot/file/1727475233899vvrorabd.jpg

Let’s go through the derivatives step by step:

### 1. \( y = \sin^2(3x - 4) \)

To find the derivative, we apply the chain rule:
\[
\frac{dy}{dx} = 2 \sin(3x - 4) \cdot \cos(3x - 4) \cdot \frac{d}{dx}(3x - 4)
\]
Since \(\frac{d}{dx}(3x - 4) = 3\):
\[
\frac{dy}{dx} = 6 \sin(3x - 4) \cdot \cos(3x - 4)
\]
Using the double angle identity for sine, this simplifies to:
\[
\frac{dy}{dx} = 3 \sin(6x - 8)
\]

### 2. \( y = \frac{\cos 2x}{3x - 4} \)

To differentiate this, we use the quotient rule. Let \( u = \cos 2x \) and \( v = 3x - 4 \). The quotient rule states:
\[
\frac{dy}{dx} = \frac{v \cdot u' - u \cdot v'}{v^2}
\]
First, find \( u' = -2 \sin 2x \) and \( v' = 3 \):
\[
\frac{dy}{dx} = \frac{(3x - 4)(-2 \sin 2x) - \cos 2x(3)}{(3x - 4)^2}
\]
Simplifying:
\[
\frac{dy}{dx} = \frac{-6x \sin 2x + 8 \sin 2x - 3 \cos 2x}{(3x - 4)^2}
\]

### 3. \( y = \frac{1}{2} \tan x \sin 2x \)

We apply the product rule:
\[
\frac{dy}{dx} = \frac{1}{2} \left[ \sec^2 x \sin 2x + \tan x \cdot 2 \cos 2x \right]
\]
This simplifies to:
\[
\frac{dy}{dx} = \frac{1}{2} \left[ \sec^2 x \sin 2x + 2 \tan x \cos 2x \right]
\]

### 4. Find \( y'' - 2y' \) if \( y = \sin 3x - \cos 3x \)

First, find the first derivative:
\[
y' = 3 \cos 3x + 3 \sin 3x
\]
Then, find the second derivative:
\[
y'' = -9 \sin 3x + 9 \cos 3x
\]
Now, calculate \( y'' - 2y' \):
\[
y'' - 2y' = (-9 \sin 3x + 9 \cos 3x) - 2(3 \cos 3x + 3 \sin 3x)
\]
\[
y'' - 2y' = -15 \sin 3x + 3 \cos 3x
\]

### 5. Solve for \( xy + 2x - 5y - y^2 = 0 \)

We use implicit differentiation:
\[
\frac{d}{dx}(xy) + \frac{d}{dx}(2x) - \frac{d}{dx}(5y) - \frac{d}{dx}(y^2) = 0
\]
The derivative of \( xy \) is:
\[
\frac{d}{dx}(xy) = x \frac{dy}{dx} + y
\]
For \( 2x \), \( \frac{d}{dx}(2x) = 2 \).
For \( 5y \), \( \frac{d}{dx}(5y) = 5 \frac{dy}{dx} \).
For \( y^2 \), \( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \).

Thus:
\[
x \frac{dy}{dx} + y + 2 - 5 \frac{dy}{dx} - 2y \frac{dy}{dx} = 0
\]

### 6. \( y = \sin^2(x^2 - x) \) using the chain rule

Using the chain rule:
\[
\frac{dy}{dx} = 2 \sin(x^2 - x) \cdot \cos(x^2 - x) \cdot \frac{d}{dx}(x^2 - x)
\]
Since \( \frac{d}{dx}(x^2 - x) = 2x - 1 \), we get:
\[
\frac{dy}{dx} = 2 \sin(x^2 - x) \cdot \cos(x^2 - x) \cdot (2x - 1)
\]

### 7. Differentiate \( xy^2 + y - x^2 = 6 \)

Using implicit differentiation:
\[
\frac{d}{dx}(xy^2) + \frac{d}{dx}(y) - \frac{d}{dx}(x^2) = 0
\]
For \( xy^2 \), apply the product rule:
\[
\frac{d}{dx}(xy^2) = x \cdot 2y \frac{dy}{dx} + y^2
\]
For \( y \), \( \frac{d}{dx}(y) = \frac{dy}{dx} \).
For \( x^2 \), \( \frac{d}{dx}(x^2) = 2x \).

Thus:
\[
x \cdot 2y \frac{dy}{dx} + y^2 + \frac{dy}{dx} - 2x = 0
\]

Would you like to go deeper into any of these steps or explanations? Here are some additional questions you might find helpful:

1. How do you apply the chain rule to trigonometric functions?
2. What is the quotient rule and how does it differ from the product rule?
3. How can implicit differentiation help when both variables depend on \(x\)?
4. What is the role of the double angle identity in simplifying trigonometric expressions?
5. Can you explain how the chain rule is used in higher-order derivatives?

**Tip**: When using implicit differentiation, always remember that the derivative of \(y\) with respect to \(x\) is \( \frac{dy}{dx} \).

Tentukan turunan dari fungsi berikut (including chain rule, product rule, quotient rule, and implicit differentiation).

This page explains the process of finding derivatives using chain rule, product rule, and quotient rule. It covers differentiation problems involving trigonometric functions and implicit differentiation, suitable for advanced high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation