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Chapter 2: Problem 14

Differentiate implicitly to find dy/dx. 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
\( \frac{dy}{dx} = \frac{12x + 5}{4y^3 + 3} \); slope at \( (1, -2) \) is \( \frac{17}{-29} \)

Step by step solution

01

Differentiate both sides with respect to x

To differentiate the equation implicitly, apply the chain rule to each term. Differentiate both sides of the equation \[ 4x^3 - y^4 - 3y + 5x + 1 = 0 \] with respect to x.
02

Differentiate each term

Differentiate each term, applying the chain rule as necessary:\[ \frac{d}{dx}(4x^3) - \frac{d}{dx}(y^4) - \frac{d}{dx}(3y) + \frac{d}{dx}(5x) + \frac{d}{dx}(1) = 0 \] This gives us \[ 12x - 4y^3 \frac{dy}{dx} - 3 \frac{dy}{dx} + 5 = 0 \]
03

Solve for \( \frac{dy}{dx} \)

Isolate the term \( \frac{dy}{dx} \). Combine the \( \frac{dy}{dx} \) terms:\[ 12x + 5 = 4y^3 \frac{dy}{dx} + 3 \frac{dy}{dx} \] Factor out \( \frac{dy}{dx} \) on the right side:\[ 12x + 5 = (4y^3 + 3) \frac{dy}{dx} \] Then solve for \( \frac{dy}{dx} \) by dividing both sides by \( 4y^3 + 3 \):\[ \frac{dy}{dx} = \frac{12x + 5}{4y^3 + 3} \]
04

Substitute the point (1, -2)

Substitute the given point \( (1, -2) \) into the derivative equation to find the slope at that point:\[ \frac{dy}{dx} \Bigg|_{(1, -2)} = \frac{12(1) + 5}{4(-2)^3 + 3} \] This simplifies to \[ \frac{dy}{dx} \Bigg|_{(1, -2)} = \frac{12 + 5}{4(-8) + 3} = \frac{17}{-32 + 3} = \frac{17}{-29} \]

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Key Concepts

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

Chain Rule
The chain rule is essential in calculus, especially when dealing with implicit differentiation. It allows us to differentiate more complex functions by breaking them into simpler parts.

When you have a composite function, where one function is nested inside another, the chain rule helps you differentiate by multiplying the derivative of the outer function by the derivative of the inner function.

For example, in the expression \( y^4 \), treating y as a function of x, we use the chain rule to get the derivative:
\[ \frac{d}{dx}(y^4) = 4y^3 \frac{dy}{dx} \].

Here, \(4y^3\) comes from differentiating \(y^4\), and \( \frac{dy}{dx} \) accounts for the inner function y. Therefore, the chain rule is useful for differentiating terms involving y implicitly. Applying it correctly ensures you do not miss critical parts of the derivative.
Implicit Differentiation
Implicit differentiation comes into play when you have an equation involving both x and y, and you can't easily solve for y as an explicit function of x.

Instead of isolating y, you differentiate both sides of the equation with respect to x, treating y as an implicit function of x.

For the exercise \[ 4x^3 - y^4 - 3y + 5x + 1 = 0 \], we differentiate term-by-term:
\[ \frac{d}{dx}(4x^3) - \frac{d}{dx}(y^4) - \frac{d}{dx}(3y) + \frac{d}{dx}(5x) + \frac{d}{dx}(1) = 0 \].

This results in: \[ 12x - 4y^3 \frac{dy}{dx} - 3 \frac{dy}{dx} + 5 = 0 \],
where each term is differentiated, and we apply the chain rule to terms involving y.

The goal is to solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation. Substitution of any given points x and y can then be done to find specific slopes at those points.
Slope of the Curve
The slope of the curve at a given point on a function tells us how steep the function is at that point. It is given by the derivative \( \frac{dy}{dx} \).

After using implicit differentiation to find \( \frac{dy}{dx} = \frac{12x + 5}{4y^3 + 3} \), we can substitute the given point \( (1, -2) \) to determine the slope at that specific location on the curve.

For example, substituting \( x = 1 \) and \( y = -2 \):
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Most popular questions from this chapter

Use your calculator's absolute-value feature to graph the follow. ing functions and determine relative extrema and intervals over which the function is increasing or decreasing. State the \(x\) -values at which the derivative does not exist. $$f(x)=\left|x^{3}-1\right|$$

The total-cost and total-revenue functions for producing \(x\) items are $$ C(x)=5000+600 x \text { and } R(x)=-\frac{1}{2} x^{2}+1000 x $$ where \(0 \leq x \leq 600 .\) a) Find the total-profit function \(P(x)\) b) Find the number of items, \(x,\) for which the total profit is a maximum.

Cost and tolerance. A painting firm contracts to paint the exterior of a large water tank in the shape of a half-dome (a hemisphere). The radius of the tank is measured to be \(100 \mathrm{ft}\) with a tolerance of 卤6 in. \((\pm 0.5 \mathrm{ft}) .\) (The formula for the surface area of a hemisphere is \(A=2 \pi r^{2} ;\) use 3.14 as an approximation for \(\pi\).) Each can of paint costs \(\$ 30\) and covers \(300 \mathrm{ft}^{2}\) a) Calculate \(d A,\) the approximate difference in the surface area due to the tolerance. b) Assuming the painters cannot bring partial cans of paint to the job, how many extra cans should they bring to cover the extra area they may encounter? c) How much extra should the painters plan to spend on paint to account for the possible extra area?

A store sells \(Q\) units of a product per year. It costs a dollars to store one unit for a year. To reorder, there is a fixed cost of \(b\) dollars, plus \(c\) dollars for each unit. How many times per year should the store reorder, and in what lot size, in order to minimize inventory costs?

Since \(1970,\) the purchasing power of the dollar, as measured by consumer prices, can be modeled by the function $$P(x)=\frac{2.632}{1+0.116 x}$$, where \(x\) is the number of years since \(1970 .\) (Source: U.S. Bureau of Economic Analysis.) a) Find $$P(10), P(20),$$ and $$P(40)$$ b) When was the purchasing power $$\$ 0.50 ?$$ c) Find $$\lim _{x \rightarrow \infty} P(x)$$. (GRAPHS CANNOT COPY)
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