Chapter 3: Problem 6
Differentiate implicily to find \(d y / d x\). Then find the slope of the curve at the given point. $$ 4 x^{3}-y^{4}-3 y+5 x+1=0 ; \quad(1,-2) $$
Short Answer
Expert verified
\( \frac{dy}{dx} = -\frac{17}{29} \)
Step by step solution
01
- Differentiate Implicitly
Given the equation \[ 4x^{3} - y^{4} - 3y + 5x + 1 = 0 \]we need to differentiate both sides with respect to x. Apply the differentiation rules: \[ \frac{d}{dx}(4x^{3}) - \frac{d}{dx}(y^{4}) - \frac{d}{dx}(3y) + \frac{d}{dx}(5x) + \frac{d}{dx}(1) = 0 \]This becomes: \[ 12x^{2} - 4y^{3}\frac{dy}{dx} - 3\frac{dy}{dx} + 5 = 0 \]
02
- Solve for \(\frac{dy}{dx}\)
Rearrange the differentiated equation to isolate \(\frac{dy}{dx}\): \[ 12x^{2} + 5 = 4y^{3}\frac{dy}{dx} + 3\frac{dy}{dx} \]Factor out \(\frac{dy}{dx}\) on the right side: \[ 12x^{2} + 5 = \left( 4y^{3} + 3 \right)\frac{dy}{dx} \]Then, solve for \(\frac{dy}{dx}\) by dividing both sides by \(4y^{3} + 3\): \[ \frac{dy}{dx} = \frac{12x^{2} + 5}{4y^{3} + 3} \]
03
- Substitute Given Point
To find the slope at the point (1, -2), substitute \(x = 1\) and \(y = -2\) into the derivative: \[ \frac{dy}{dx} \Bigg|_{(1,-2)} = \frac{12(1)^{2} + 5}{4(-2)^{3} + 3} \]Calculate the values: \[ \frac{dy}{dx} \Bigg|_{(1,-2)} = \frac{12 + 5}{-32 + 3} = \frac{17}{-29} = -\frac{17}{29} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
Derivatives measure how a function changes as its input changes. It represents the rate at which one quantity changes with respect to another. For example, if you have a function f(x), the derivative, denoted as f'(x) or \( \frac{df}{dx} \), tells you how f(x) changes as x changes. In simple terms, it gives you the slope of the function at any given point.
To compute the derivative of a sum or difference of functions, apply the rule: \( (f(x) + g(x))' = f'(x) + g'(x) \). And for a product or quotient of functions, use the product rule: \( (fg)' = f'g + fg' \), and quotient rule: \( \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \).
Derivatives play a crucial role in fields like physics, engineering, economics, and beyond, as they help to model and predict dynamic systems.
To compute the derivative of a sum or difference of functions, apply the rule: \( (f(x) + g(x))' = f'(x) + g'(x) \). And for a product or quotient of functions, use the product rule: \( (fg)' = f'g + fg' \), and quotient rule: \( \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \).
Derivatives play a crucial role in fields like physics, engineering, economics, and beyond, as they help to model and predict dynamic systems.
Implicit Functions
Implicit functions are those where the dependent variable y is not isolated on one side of the equation. Instead, y and x are intertwined. For example, in the equation \( 4x^3 - y^4 - 3y + 5x + 1 = 0 \), y is not explicitly solved for in terms of x.
Implicit differentiation is used when it’s difficult or impossible to solve the function for y explicitly. To differentiate implicitly:
By differentiating implicitly, we can find the slope of the curve even when y is not isolated.
Implicit differentiation is used when it’s difficult or impossible to solve the function for y explicitly. To differentiate implicitly:
- Differentiate both sides of the equation with respect to x.
- Apply the chain rule to terms involving y, remembering that y is a function of x.
- Rearrange the resulting equation to solve for \( \frac{dy}{dx} \).
By differentiating implicitly, we can find the slope of the curve even when y is not isolated.
Slope of a Curve
The slope of a curve at any point indicates how steep the curve is at that particular location. Mathematically, it is represented by the derivative \( \frac{dy}{dx} \).
To find the slope at a specific point, say (1, -2), after finding the derivative, substitute the values of x and y into the derivative equation.
For example, given the implicit function differential result \( \frac{dy}{dx} = \frac{12x^2 + 5}{4y^3 + 3} \), to find the slope at the point (1, -2), substitute x = 1 and y = -2:
\[ \frac{dy}{dx} \Bigg|_{(1,-2)} = \frac{12 (1)^2 + 5}{4 (-2)^3 + 3} = \frac{17}{-29} = -\frac{17}{29} \]
This gives the slope of the curve at point (1, -2).
To find the slope at a specific point, say (1, -2), after finding the derivative, substitute the values of x and y into the derivative equation.
For example, given the implicit function differential result \( \frac{dy}{dx} = \frac{12x^2 + 5}{4y^3 + 3} \), to find the slope at the point (1, -2), substitute x = 1 and y = -2:
\[ \frac{dy}{dx} \Bigg|_{(1,-2)} = \frac{12 (1)^2 + 5}{4 (-2)^3 + 3} = \frac{17}{-29} = -\frac{17}{29} \]
This gives the slope of the curve at point (1, -2).
Differentiation Rules
Differentiation rules help simplify the process of finding derivatives. The basic rules include:
In implicit differentiation, use the chain rule for y terms, treating y as a function of x. For example, differentiating \( y^4 \) with respect to x gives \( 4y^3 \frac{dy}{dx} \).
These rules make the process more straightforward and reduce errors.
- Power Rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \)
- Constant Rule: \( \frac{d}{dx}(c) = 0 \)
- Sum Rule: \( \frac{d}{dx}(f+g) = f' + g' \)
- Product Rule: \( \frac{d}{dx}(fg) = f'g + fg' \)
- Quotient Rule: \( \frac{d}{dx}(\frac{f}{g}) = \frac{f'g - fg'}{g^2} \)
- Chain Rule: \( \frac{d}{dx}(f(g(x))) = f'(g(x))g'(x) \)
In implicit differentiation, use the chain rule for y terms, treating y as a function of x. For example, differentiating \( y^4 \) with respect to x gives \( 4y^3 \frac{dy}{dx} \).
These rules make the process more straightforward and reduce errors.