Question

Carry out the following steps. a. Use implicit differentiation to find \(\frac{d y}{d x}\). b. Find the slope of the curve at the given point. $$\frac{x}{y^{2}+1}=1 ;(10,3)$$

Short answer

Answer: The slope of the curve at the point (10, 3) is $\frac{3}{2}$.

Step-by-step solution

Rewrite the equation in the standard form

Before performing implicit differentiation, let's first rewrite the given equation in the standard form so it's easier to work with: $$\frac{x}{y^2+1}-1=0$$

Apply implicit differentiation

Now we need to find \(\frac{dy}{dx}\) by implicitly differentiating the equation with respect to \(x\). To do this, we'll differentiate both sides of the equation with respect to \(x\). For the left side, we'll differentiate term by term: The derivative of \(x\) with respect to \(x\) is 1: $$\frac{d}{dx}\left(\frac{x}{y^2+1}\right) = 1$$ Now we need to differentiate \(\frac{x}{y^2+1}\) with respect to \(x\). We apply the quotient rule, which states \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}\), where \(u=x\) and \(v=y^2+1\). In this case, \(u'=1\) and \(v'=\frac{d}{dx}(y^2+1)=2y\frac{dy}{dx}\), since we need to use the chain rule for differentiating \(y^2\). Plugging in these values into the quotient rule formula, we get: $$\frac{d}{dx}\left(\frac{x}{y^2+1}\right) = \frac{(y^2+1)(1) - x(2y\frac{dy}{dx})}{(y^2+1)^2} = 1$$ Now we have: $$\frac{(y^2+1) - x(2y\frac{dy}{dx})}{(y^2+1)^2} = 1$$

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

To find the expression for \(\frac{dy}{dx}\), we need to solve the above equation for \(\frac{dy}{dx}\). First, we'll multiply both sides by \((y^2+1)^2\) to get rid of the denominator: $$(y^2+1) - x(2y\frac{dy}{dx})=(y^2+1)^2$$ Now, let's isolate the term containing \(\frac{dy}{dx}\): $$x(2y\frac{dy}{dx}) = (y^2+1)^2-(y^2+1)$$ Next, we'll divide by \(x(2y)\) to solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = \frac{(y^2+1)^2-(y^2+1)}{x(2y)}$$

Find the slope at the given point

Now, we'll evaluate the expression for \(\frac{dy}{dx}\) at the given point \((10,3)\): $$\frac{dy}{dx}(10,3) = \frac{(3^2+1)^2-(3^2+1)}{10(2\cdot3)}$$ Calculating the values, we get: $$\frac{dy}{dx}(10,3) = \frac{(10)^2-10}{60}$$ $$\frac{dy}{dx}(10,3) = \frac{90}{60}$$ $$\frac{dy}{dx}(10,3) = \frac{3}{2}$$ The slope of the curve at the given point \((10, 3)\) is \(\frac{3}{2}\).

Key concepts

Quotient Rule

The quotient rule is a powerful tool when dealing with differentiation of functions written as a fraction. In mathematics, the ability to differentiate effectively is key to solving many problems involving rates of change. When you're given a function where one expression is divided by another, such as \(\frac{u}{v}\), the quotient rule comes into play. It's formally expressed as \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}\), where \(u\) and \(v\) are functions of \(x\), and \(u'\) and \(v'\) are their respective derivatives.

Applying the quotient rule requires you to identify \(u\) and \(v\) within the function you’re differentiating. Once identified, compute their derivatives separately. Then, apply the quotient formula by multiplying \(v\) with the derivative of \(u\) (\(u'\)), and \(u\) with the derivative of \(v\) (\(v'\)), subtracting the second product from the first, and finally dividing the result by \(v^2\). Remember, attention to detail is crucial here because a mistake in differentiating \(u\) or \(v\) can lead to an entire miscalculation of the derivative.

For the implicit differentiation problem provided, identifying \(u = x\) and \(v = y^2 + 1\) was the first step. The derivative of \(x\) with respect to \(x\) is 1, so \(u' = 1\), and since \(v\) is a function of \(y\), which in turn is a function of \(x\), we need to also apply the chain rule to find \(v'\).

Chain Rule

The chain rule is a formula to compute the derivative of a composition of functions. If you have a function \(h(x)\) that can be expressed as the composition of two functions, \(f(g(x))\), the derivative of \(h\) with respect to \(x\) is the derivative of \(f\), evaluated at \(g(x)\), multiplied by the derivative of \(g\) with respect to \(x\). In mathematical terms, \(\frac{d}{dx}h(x) = f'(g(x)) \times g'(x)\).

The beauty of the chain rule is its ability to break down the differentiation of complex expressions into simpler parts. In the case of our problem, where the function \(v = y^2 + 1\) depends on \(y\), and \(y\) depends on \(x\), the chain rule allows us to differentiate \(v\) with respect to \(x\) by first taking the derivative of \(v\) with respect to \(y\), \(v' = 2y\), and then multiplying it by the derivative of \(y\) with respect to \(x\), \(\frac{dy}{dx}\). This intermediate step is crucial in correctly applying the quotient rule as it accounts for the dependency of \(y\) on \(x\).

Understanding and applying the chain rule correctly is essential when differentiating complex functions, especially when they involve implicit differentiation where one variable is a function of another as seen in our exercise.

Slope of a Curve

The slope of a curve at a particular point is the rate at which the \(y\) value changes with respect to the \(x\) value, or in other words, it is the derivative of the function at that point. In more familiar terms, it tells us how steep the curve is at a particular point. The slope is often denoted by \(\frac{dy}{dx}\) when dealing with continuous functions.

When you're given a curve defined by an equation, finding the slope at a particular point requires you to first find its derivative with respect to \(x\). This derivative represents the slope of the tangent line to the curve at any point \(x\), and evaluating this derivative at a specific point gives us the slope at that point. For instance, in the exercise provided, after carrying out implicit differentiation, finding the derivative \(\frac{dy}{dx}\) gives us a general expression for the slope of the curve. Substituting the given point \(10, 3\) into this expression allows us to compute the slope of the curve at that exact location.

The slope can be thought of as a snapshot of the curve's behavior at a particular point, and it plays a critical role in understanding the nature of the curve. For example, a positive slope indicates that the curve is rising as it moves from left to right, while a negative slope suggests that the curve is falling. If the slope is zero, it indicates a flat, horizontal tangent, implying that the curve has reached a local maximum or minimum at that point.