Question

Determine whether the following statements are true and give an explanation or counterexample. a. For any equation containing the variables $x$ and $y,$ the derivative $dy/dx$ can be found by first using algebra to rewrite the equation in the form $y=f(x).$ b. For the equation of a circle of radius $r,x^2+y^2=r^2,$ we have $\frac{dy}{dx}=-\frac{x}{y},$ for $y\ne 0$ and any real number $r>0.$ c. If $x=1$, then by implicit differentiation, $1=0.$ d. If $xy=1,$ then $y^{\prime }=1/x.$

Short answer

Question: Determine if the given statements are true or false, and provide explanations or counterexamples. a. Rewriting the equation in the form y = f(x) can always find the derivative dy/dx. b. For the equation of a circle with any radius r > 0, given by $x^2+y^2=r^2$ and $y\ne 0$, it is true that $\frac{dy}{dx}=-\frac{x}{y}$. c. If x = 1, then by implicit differentiation, 1 = 0. d. For the equation xy = 1, it is true that $y^{\prime }=\frac{1}{x}$. Answer: a. False Explanation: Derivative dy/dx can be found using implicit differentiation for the cases where the equation cannot be rewritten in the explicit form y = f(x). b. True Explanation: By using implicit differentiation on the circle equation, we derived the equation $\frac{dy}{dx}=-\frac{x}{y}$. c. False Explanation: The claim doesn't make sense without the context of an equation, and we cannot jump to a conclusion without implicit differentiation applied to a specific equation. d. False Explanation: Using implicit differentiation on the given equation, we derived the expression for the derivative as $\frac{dy}{dx}=-\frac{1}{x^2}$, which is different from the given statement.

Step-by-step solution

a. Determining if rewriting the equation in the form y = f(x) can find the derivative dy/dx

Theorem: For any differentiable function y = f(x), its derivative dy/dx exists and can be found by the differentiation rules. However, not every equation containing variables x and y can be rewritten in the explicit form y = f(x). Some equations can be implicitly differentiated to find dy/dx. So, this statement is false. Counterexample: Consider the equation of a circle $x^2+y^2=r^2$. We can't isolate y for a circle, as there will always be two values of y for a single value of x, except at the points $(\pm r,0)$. So, there is no corresponding explicit function f(x) for a circle.

b. Deriving the equation of a circle

To find whether $\frac{dy}{dx}=-\frac{x}{y}$ for the equation of a circle with any radius r > 0, given by $x^2+y^2=r^2$ and $y\ne 0$, we will use implicit differentiation. Differentiate both sides of the equation with respect to x: $$\frac{d}{dx}(x^2+y^2)=\frac{d}{dx}(r^2)$$ $$2x+2y\frac{dy}{dx}=0$$ Now solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=-\frac{x}{y}$$ This shows that the statement is true, and we have derived the required equation for the circle.

c. Evaluating x=1 with implicit differentiation

The claim that if x = 1 then by implicit differentiation, 1 = 0, doesn't make sense as it is not derived from any given equation. For any equation, it is necessary to use implicit differentiation to find the derivative and then evaluate it at the specified point. We cannot jump to a conclusion without the context of an equation.

d. Solving the equation xy = 1

Given the equation xy = 1, we need to determine if $y^{\prime }=\frac{1}{x}$. First, implicitly differentiate both sides of the equation with respect to x: $$\frac{d}{dx}(xy)=\frac{d}{dx}(1)$$ Using the product rule, we get: $$(1)\frac{dy}{dx}+(y)(1)=0$$ Now solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=-y$$ Since $xy=1$, we have $y=\frac{1}{x}$, which implies $\frac{dy}{dx}=-\frac{1}{x^2}$. The given statement is false, as our derived expression for the derivative is different from the given one. In conclusion, we have examined the given statements, determining whether they are true or false, and provided explanations or counterexamples for our conclusions.

Key concepts

Derivative

In calculus, a derivative is a fundamental concept that represents how a function changes as its input changes. It's like measuring how fast something is happening or the slope of a curve at any given point. Let's break it down further:

- **Instantaneous Rate of Change**: Think of it as how fast something is moving at an exact moment, similar to looking at your speedometer while driving.
- **Slope of a Curve**: On a graph, a derivative tells you the steepness of the curve at any given point.The derivative of a function is typically represented as $\frac{dy}{dx}$, meaning the rate of change of $y$ with respect to $x$. When dealing with complex equations, implicit differentiation becomes a useful tool to find this rate of change without rewriting the equation in specific forms.

Equation of a Circle

An equation of a circle is a neat way to describe all the points that make up the circle. The standard form of an equation of a circle is $x^2+y^2=r^2$, where $r$ is the radius. Here’s how this works:

- **All Points the Same Distance from Center**: Every point $(x,y)$ on the circle is exactly $r$ units from the center, typically the origin $(0,0)$.
- **Cannot Isolate $y$**: Unlike some equations, it's difficult to rewrite this as $y=f(x)$, because for each $x$, there are usually two corresponding $y$ values.Implicit differentiation becomes crucial here, as it allows us to find derivatives, like $\frac{dy}{dx}=-\frac{x}{y}$, without needing to solve for $y$ explicitly.

Product Rule

The product rule is a vital differentiation rule that helps us find the derivative of a product of two functions. Let’s look at how it works:

- **Formula**: When you have two functions $u(x)$ and $v(x)$, the derivative of their product $uv$ is $u^{\prime }v+u{v}^{\prime }$.
- **Application**: You apply the rule when differentiating expressions like $xy=1$. Here, you don't separate $x$ and $y$ but treat them as parts of a single product.

The product rule makes it straightforward to handle equations involving products of functions, especially when using implicit differentiation.

Differentiation Rules

Differentiation rules are guidelines that help us find derivatives successfully and efficiently. These rules include ways to handle different mathematical expressions. Here are some basics:

- **Power Rule**: Useful for monomials, if $f(x)=x^n$, then $f^{\prime }(x)=n{x}^{n-1}$.
- **Sum Rule**: The derivative of a sum $(f+g)$ is just the sum of the derivatives $f^{\prime }+g^{\prime }$.
- **Product and Quotient Rules**: These help differentiate products and quotients of functions and are often used in tandem with implicit differentiation.

Differentiation rules simplify complex problems, making it easier to find derivatives without needing to constantly break everything down into simpler expressions.