Question

Tangent lines Find an equation of the line tangent to the graph of \(f\) at the given point. $$f(x)=\cos ^{-1} x^{2} ;(1 / \sqrt{2}, \pi / 3)$$

Short answer

Question: Determine the equation of the tangent line to the graph of the function \(f(x) = \cos^{-1}(x^2)\) at the point \((1/\sqrt{2}, \pi/3)\). Answer: The equation of the tangent line is \(y = -\sqrt{2}x + \sqrt{2} + \frac{\pi}{3}\).

Step-by-step solution

Differentiate the Function

First, we'll find the derivative of the function \(f(x) = \cos^{-1}(x^2)\). To do this, we'll use the chain rule: \(\frac{d}{dx}[\cos^{-1}(u)] = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}\). Let \(u=x^2\), then \(\frac{du}{dx} = 2x\). So, $$f'(x) = -\frac{1}{\sqrt{1-x^4}} \cdot 2x$$

Evaluate the Derivative at the Given Point

Next, we'll find the slope of the tangent line by evaluating the derivative at \(x = 1/\sqrt{2}\): $$f'\left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^4}} \cdot 2\left(\frac{1}{\sqrt{2}}\right) = -\frac{\sqrt{2}}{\sqrt{1-\frac{1}{2}}} = -\sqrt{2}$$ This implies that the slope of the tangent line at the given point is \(- \sqrt{2}\).

Find the Equation of the Tangent Line

Now, we'll use the point-slope form of a line to find the equation of the tangent line: \(y-y_0 = m(x-x_0)\), where \((x_0, y_0)\) is the given point and \(m\) is the slope of the tangent line. Substituting the values, we get: $$y - \frac{\pi}{3} = -\sqrt{2} \left(x - \frac{1}{\sqrt{2}}\right)$$ Expanding and simplifying the equation, we get: $$y = -\sqrt{2}x + \sqrt{2} + \frac{\pi}{3}$$ Therefore, the equation of the tangent line is \(y = -\sqrt{2}x + \sqrt{2} + \frac{\pi}{3}\).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus, which involves finding the rate at which a function changes at any given point. It is crucial when solving problems involving rates of change, and it helps us determine the slopes of curves, among other applications.
In the exercise provided, we need to differentiate the function \( f(x) = \cos^{-1}(x^2) \). Differentiation allows us to find the derivative \( f'(x) \), which represents the slope of the tangent line to the curve at any point \( x \).
To carry out differentiation, various rules and techniques can be utilized. These include basic differentiation rules, such as the power rule, and more advanced techniques, like the chain rule, which is particularly useful when dealing with composite functions.

Chain Rule

The chain rule is one of the foundational techniques in calculus used to differentiate composite functions. When a function is composed of two or more nested functions, the chain rule becomes essential to find the derivative accurately.
In our exercise, we differentiate \( f(x) = \cos^{-1}(x^2) \) using the chain rule. Here, we consider \( u = x^2 \) as an inner function and \( \cos^{-1}(u) \) as an outer function. Applying the chain rule involves differentiating the outer function with respect to the inner function and then multiplying it by the derivative of the inner function.
The general formula for the chain rule is:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

The complete differentiation step:

• Differentiate \( \cos^{-1}(u) \) and get \( -\frac{1}{\sqrt{1-u^2}} \).
• Then differentiate \( u = x^2 \) to get \( 2x \).
• Finally, multiply the results: \( f'(x) = -\frac{1}{\sqrt{1-x^4}} \cdot 2x \).

This illustrates how the chain rule helps unravel and differentiate complex functions efficiently.

Point-Slope Form

The point-slope form is a helpful tool in finding the equation of a line when you have a point on the line and its slope. This form is expressed as \( y - y_0 = m(x - x_0) \), where \((x_0, y_0)\) is a known point, and \( m \) represents the slope.
In our problem, after calculating the derivative, we determined the slope of the tangent line at the point \( (\frac{1}{\sqrt{2}}, \frac{\pi}{3}) \) to be \(-\sqrt{2}\). Using the point-slope formula, we replaced \( m \) with \(-\sqrt{2}\), \( x_0 \) with \( \frac{1}{\sqrt{2}} \), and \( y_0 \) with \( \frac{\pi}{3} \).
This allows us to write the equation of the tangent line as:

• \( y - \frac{\pi}{3} = -\sqrt{2} (x - \frac{1}{\sqrt{2}}) \)

To express it in slope-intercept form, you can rearrange and simplify the equation:

• Expanding gives \( y = -\sqrt{2}x + \sqrt{2} + \frac{\pi}{3} \).

This tangent line equation tells you how the line touches the curve precisely at the given point.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find angles when given their trigonometric values. They are the inverse operations of the standard trigonometric functions like sine, cosine, and tangent.
In the exercise, the function \( f(x) = \cos^{-1}(x^2) \) indicates the inverse cosine function applied to \( x^2 \). The inverse cosine, represented as \( \cos^{-1}(x) \), answers the question: "What angle has a cosine of \( x \)?"
These functions have specific derivatives, which are crucial for problems involving calculus. The derivative of \( \cos^{-1}(u) \) is \( -\frac{1}{\sqrt{1-u^2}} \), which we used to differentiate our function by coupling it with the chain rule.
Understanding inverse trigonometric functions is essential for differentiating and integrating them in calculus problems, especially when working with composite functions and finding derivatives as we did in this exercise.