Problem 7 Find a formula for the derivativ... [FREE SOLUTION]

Chapter 4: Problem 7

Find a formula for the derivative $y^{\prime}$ at the point $(x, y):$ $$ \sin (x+y)=x y \Rightarrow $$

Short Answer

Expert verified

The derivative formula is $ y' = \frac{y - \cos(x+y)}{\cos(x+y) - x} $.

Step by step solution

Differentiate Both Sides Implicitly

We start with the given equation $ \sin(x+y) = xy $. To find the derivative $ y' $, we differentiate both sides with respect to $ x $. The derivative of $ \sin(x+y) $ is $ \cos(x+y) \cdot (1 + y') $ because of the chain rule: the derivative of $ x+y $ with respect to $ x $ is $ 1 + y' $. The derivative of $ xy $ is $ x y' + y $ due to the product rule. Thus, the differentiated equation is:\[ \cos(x+y) \cdot (1 + y') = x y' + y \]

Solve for y-prime (y')

Next, we isolate $ y' $ in the derived equation. Starting with:\[ \cos(x+y) \cdot (1 + y') = x y' + y \]Expand the left side:\[ \cos(x+y) + \cos(x+y) y' = x y' + y \]Rearrange to group all terms containing $ y' $ on the left side:\[ \cos(x+y) y' - x y' = y - \cos(x+y) \]Factor out $ y' $:\[ y'(\cos(x+y) - x) = y - \cos(x+y) \]Solve for $ y' $:\[ y' = \frac{y - \cos(x+y)}{\cos(x+y) - x} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule

To understand the chain rule, imagine it as a technique for finding the derivative of composite functions. It is essential when you have a function inside of another function. For example, with the function $ \sin(x+y) $, you are dealing with $ \sin(u) $ where $ u = x+y $.
This means we have to differentiate $ \sin(u) $ first, and then multiply it by the derivative of $ u $ with respect to $ x $.
The chain rule generalizes to $ f(g(x)) $ formats, written as $ f'(g(x)) \cdot g'(x) $.
\[ \text{If } f(x) = \sin(x+y), \text{ the derivative is } \cos(x+y) \cdot (1 + y') \]

The $ \cos(x+y) $ results from differentiating $ \sin(u) $.
The $ 1+y' $ comes from the derivative of $ u=x+y $.

Use the chain rule whenever derivatives of nested functions come into play.

Product Rule

The product rule comes into use when differentiating a product of two functions. It states that for two functions $ u(x) $ and $ v(x) $, the derivative of their product $ u(x)v(x) $ is:
\[ u'(x)v(x) + u(x)v'(x) \]
In the given problem, the expression $ xy $ consists of two functions: $ x $ and $ y $.

The derivative formula becomes:
\[ \frac{d}{dx}(xy) = x \cdot \frac{dy}{dx} + y \cdot \frac{d}{dx}(x) \]

For $ x \cdot y' $, treat $ y $ as a function of $ x $ while $ x $ remains as it is.
For $ y $, since the derivative of $ x $ with respect to itself is 1, it directly contributes $ y $.

The product rule simplifies finding the derivative of function products, often seen in implicit differentiation situations.

Differentiation

Differentiation is the foundational method in calculus used to find the rate of change of a function. It's the process of computing a derivative, which tells us how a function changes as its input changes.
Here, we differentiate the implicit function $ \sin(x+y) = xy $ with respect to $ x $ to find $ y' $. Differentiation involves applying different rules like the chain rule and product rule to break down complex functions.

Implicit differentiation lets us differentiate without solving for $ y $. It treats $ y $ as a function of $ x $ without explicit expression.
This means both sides of the equation $ \sin(x+y) = xy $ are differentiated with respect to $ x $, leading to a relation that includes $ y' $.

Mastering differentiation equips you with tools to handle a wide range of real-world problems where rates of change are studied.

Calculus Problem Solving

Solving calculus problems involves a strategic approach to applying the correct concepts and rules. In this problem, you begin with the equation $ \sin(x+y) = xy $ and differentiate it to find $ y' $. You need to identify which differentiation rules are necessary based on the forms present.

The solution process involves:

Identifying that both chain and product rules are needed.
Applying the chain rule to $ \sin(x+y) $ and the product rule to $ xy $.
Rearranging the differentiated equation to isolate $ y' $.

This approach requires recognizing the problem structure and using differentiation rules effectively.
Break down each step, focus on applying rules correctly, and rearrange your terms to solve for the desired derivative. Calculus problem-solving translates into systematic thinking and precision when dealing with changes.

91影视

Find a formula for the derivative \(y^{\prime}\) at the point \((x, y):\) $$ \sin (x+y)=x y \Rightarrow $$

Short Answer

Step by step solution

Differentiate Both Sides Implicitly

Solve for y-prime (y')

Key Concepts

Chain Rule

Product Rule

Differentiation

Calculus Problem Solving

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