Problem 62 - Calculus of a Single Variable

Question

Find an equation of the tangent line to the graph of the function at the given point. \(y=\operatorname{arcsec} 4 x, \quad\left(\frac{\sqrt{2}}{4}, \frac{\pi}{4}\right)\)

Step-by-Step Solution

Answer

The equation of the tangent line to the given function at the given point is \(y=\frac{\pi}{4}+\sqrt{2}(x-\frac{\sqrt{2}}{4}).\)

Step 1: Find the derivative of the function

The derivative of \(y=\operatorname{arcsec} 4x\) can be obtained using the chain rule and the fact that \(\frac{d}{dx}(\operatorname{arcsec} u)= \frac{1}{|u|\sqrt{u^2-1}}\) for \(|u|>1\). So, the derivative will be \(\frac{d}{dx}(\operatorname{arcsec} 4x)= \frac{1}{|4x|\sqrt{(4x)^2-1}}\cdot4.\)

Step 2: Substitute the x-coordinate into the derivative

Substitute \(x=\frac{\sqrt{2}}{4}\) into the derivative to get the slope of the tangent line at that point.

Step 3: Substitute into the equation of the line

Now substitute the slope and the coordinates of the point \((\frac{\sqrt{2}}{4}, \frac{\pi}{4})\) into the equation of the tangent line \(y-y_1=m(x-x_1)\) to get the equation of the line.

Key Concepts

Arcsecant Function

The arcsecant function is the inverse of the secant function. This means it reverses what the secant function does. If you take the secant of an angle, then apply the arcsecant to that result, you will return to the original angle.
It's important to note that the arcsecant function is defined only for values

• that are equal to or greater than 1
• and for values less than or equal to -1.

This restriction ensures the functionality of its inverse nature. In a given function like \(y = \operatorname{arcsec} 4x\), you're taking this inverse with a transformation factor of 4 on \(x\).
As you work with arcsecant, remember that it's not as smooth around 0, and requires careful handling particularly in calculus operations like differentiation.

Derivative

The derivative of a function measures how the function changes as its input changes. In simpler terms, it gives the slope of the tangent line to the $function at any given point. For the arcsecant function, the formula for the derivative is \(\frac{d}{dx}(\operatorname{arcsec} u)= \frac{1}{|u|\sqrt{u^2-1}}\).When a function has a more complex form like \(y = \operatorname{arcsec} 4x\), you'll use the chain rule to differentiate it, which we'll explain more in the next section. Understanding derivatives is crucial as it forms the basis for finding tangent lines and analyzing functions."

Chain Rule

The chain rule is a fundamental technique in calculus used when dealing with composite functions. A composite function is when one function is inside another, like \(y = \operatorname{arcsec} 4x\). Here, \(4x\) is inside the arcsecant function.The chain rule helps us differentiate such functions by taking the derivative of the outer function, evaluated at the inner function, and then multiplying it by the derivative of the inner function. So, for \(\operatorname{arcsec} 4x\), while following the formula for \(\operatorname{arcsec} u\), you multiply by the derivative of \(4x\). Thus, your derivative calculation involves two parts: the outer derivative from the arcsecant, and the inner derivative which is \(4\) from the term \(4x\). Understanding the chain rule is key to handling more complex derivatives accurately.

Equation of a Line

To find the tangent line to a curve at a given point, you need the line's slope and a point it passes through. The equation of a line typically follows the format \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line.Given the point \((\frac{\sqrt{2}}{4}, \frac{\pi}{4})\) on the curve of \(y=\operatorname{arcsec} 4x\), the derivative at this point gives the slope of the tangent line. Once you have this slope, replace \(m\), \(x_1\), and \(y_1\) in the formula to find the specific equation of the tangent line.This calculation links the geometric idea of 'tangency' with the algebraic form of a line, bridging two key areas of mathematics into a useful tool.