Problem 50 - Calculus - AP Edition

Question

Find an equation for the tangent line to the graph at the specified value of \(x\) $$y=\frac{x}{\sqrt{1-x^{2}}}, x=0$$

Step-by-Step Solution

Answer

The equation of the tangent line at \(x = 0\) is \(y = x\).

Step 1: Identify the Function and Point

The given function is \(y = \frac{x}{\sqrt{1-x^{2}}}\). We need to find the tangent line at \(x = 0\).

Step 2: Find the Value of the Function at the Point

Substitute \(x = 0\) into the function to find \(y\). \[ y = \frac{0}{\sqrt{1-0^{2}}} = 0 \]. So the point on the graph is \((0, 0)\).

Step 3: Find the Derivative of the Function

To find the slope of the tangent line, we need the derivative \( y' \) of the function.The function is \( y = \frac{x}{\sqrt{1-x^2}} = x(1-x^2)^{-1/2} \).Using the product rule: if \( u = x \) and \( v = (1-x^2)^{-1/2} \), then:\( y' = u'v + uv' \).

Step 4: Apply the Product Rule to Find Derivative

For \( u = x \), we have \( u' = 1 \).For \( v = (1-x^2)^{-1/2} \), we first find \( v' \) using the chain rule:\( v' = -\frac{1}{2}(1-x^2)^{-3/2}(-2x) = \frac{x}{(1-x^2)^{3/2}} \).Substitute into the product rule:\( y' = 1 \cdot (1-x^2)^{-1/2} + x \cdot \frac{x}{(1-x^2)^{3/2}} \).Simplify:\( y' = (1-x^2)^{-1/2} + \frac{x^2}{(1-x^2)^{3/2}} \).

Step 5: Evaluate the Derivative at the Point

Substitute \(x = 0\) into the derivative \( y'\):\( y'(0) = (1-0^2)^{-1/2} + \frac{0^2}{(1-0^2)^{3/2}} = 1 \).So the slope of the tangent line at \(x = 0\) is \(1\).

Step 6: Write the Equation of the Tangent Line

The equation of a line is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope.Using the point \((0, 0)\) and \(m = 1\), we have:\( y - 0 = 1(x - 0) \).Thus, the equation of the tangent line is: \( y = x \).

Key Concepts

Derivative

When studying calculus, the concept of a derivative is fundamental. A derivative represents the slope of the tangent line to a function at a given point. It tells us how quickly the function is changing at that point. You can think of it as the rate of change or the gradient of the function. The process of finding a derivative is called differentiation. Differentiation is a method to determine the rate at which a quantity changes. In this exercise, we needed the derivative of the function \( y = \frac{x}{\sqrt{1-x^2}} \) to find the slope of the tangent line at \( x = 0 \). Once we have the derivative, it can be evaluated at the specific point to find the desired slope. This slope is then used to write the equation for the tangent line.

Product Rule

The product rule is an essential technique in calculus, particularly useful when differentiating expressions where two functions are multiplied together. The rule states that to differentiate the product of two functions, say \( u(x) \) and \( v(x) \), you differentiate both functions separately and then apply:

• \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

In the exercise, the function \( y = \frac{x}{\sqrt{1-x^2}} \) could be rewritten as \( y = x(1-x^2)^{-1/2} \), allowing us to apply the product rule. Here, we identified \( u = x \) and \( v = (1-x^2)^{-1/2} \). The derivative of \( u \) is straightforward: \( u' = 1 \), while \( v' \) required further differentiation using another rule known as the chain rule. By putting both derivatives into the product rule, we obtain the derivative of the product which helps in finding the slope for the tangent line.

Chain Rule

The chain rule is another crucial aspect of differentiation, making it easier to handle composite functions—those composed of two or more functions. It states that if you have a composite function \( y(f(x)) \), its derivative could be found using:

• \( \frac{dy}{dx} = \frac{dy}{df} \cdot \frac{df}{dx} \)

In our exercise, the function \( (1-x^2)^{-1/2} \) required the use of the chain rule to find its derivative. First, recognize the outer function as \( g(t) = t^{-1/2} \) and the inner function as \( t(x) = 1-x^2 \). Differentiating these separately results in \( g'(t) = -\frac{1}{2}t^{-3/2} \) and \( t'(x) = -2x \). By applying the chain rule, we calculated the derivative \( v' = \frac{x}{(1-x^2)^{3/2}} \). This shows the power of the chain rule, providing us with the necessary component to utilize in tackling derivatives of more complex expressions, as seen in combination with the product rule.