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Determine an equation for the tangent to each function at the point with the given \(x\) -coordinate. a. \(f(x)=\tan x, x=\frac{\pi}{4}\) b. \(f(x)=6 \tan x-\tan 2 x, x=0\)

Short Answer

Expert verified
a. The equation for the tangent is \(y - 1 = 2\left(x - \frac{\pi}{4}\right)\). b. The equation for the tangent is \(y = 4x\).

Step by step solution

01

Differentiate the function for part (a)

We need to find the derivative of the function \(f(x) = \tan x\), as the derivative gives the slope of the tangent at any point \(x\). The derivative of \(\tan x\) with respect to \(x\) is \(\sec^2 x\). Thus, \(f'(x) = \sec^2 x\).
02

Evaluate the derivative at the given point for part (a)

To find the slope of the tangent at the given \(x\)-coordinate, we substitute \(x = \frac{\pi}{4}\) into the derivative. Thus, \(f'\left(\frac{\pi}{4}\right) = \sec^2\left(\frac{\pi}{4}\right)\). We know \(\sec\left(\frac{\pi}{4}\right) = \sqrt{2}\), so \(f'\left(\frac{\pi}{4}\right) = (\sqrt{2})^2 = 2\).
03

Determine the point on the curve for part (a)

Substitute \(x = \frac{\pi}{4}\) into the original function \(f(x) = \tan x\) to find the \(y\)-coordinate. Hence, \(f\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1\). Thus, the point is \(\left(\frac{\pi}{4}, 1 \right)\).
04

Write the equation of the tangent line for part (a)

Using the point-slope form \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point on the tangent (\(\frac{\pi}{4}, 1\)) and \(m = 2\), substitute to get \(y - 1 = 2(x - \frac{\pi}{4})\).
05

Differentiate the function for part (b)

For \(f(x) = 6\tan x - \tan 2x\), we find \(f'(x)\) using the derivatives: \(\frac{d}{dx}(6\tan x) = 6\sec^2 x\) and \(\frac{d}{dx}(\tan 2x) = 2\sec^2 2x\). Therefore, \(f'(x) = 6\sec^2 x - 2\sec^2 2x\).
06

Evaluate the derivative at the given point for part (b)

Substitute \(x = 0\) into the derivative. Compute \(f'(0) = 6\sec^2(0) - 2\sec^2(0)\), knowing \(\sec(0) = 1\). Therefore, \(f'(0) = 6(1) - 2(1) = 4\).
07

Determine the point on the curve for part (b)

Substitute \(x = 0\) into the original function \(f(x) = 6\tan x - \tan 2x\). Since \(\tan(0) = 0\), \(f(0) = 6(0) - 0 = 0\). Thus, the point is \((0, 0)\).
08

Write the equation of the tangent line for part (b)

Using the point-slope form \(y - y_1 = m(x - x_1)\), substitute the point \((0, 0)\) and the slope \(4\). We get \(y = 4x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a core concept in calculus focused on finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point. This is especially useful for determining the slope of a tangent line to a curve, which is a straight line that touches a curve at one point without crossing it. To differentiate a function, we apply rules like the power rule, product rule, and in the case of trigonometric functions such as the tangent function, the chain and trigonometric rules. Remember, the derivative of \( \tan x \) with respect to \( x \) is \( \sec^2 x \). Differentiation helps in many fields, from physics to engineering, by analyzing how small changes in input affect outputs. In the exercise, differentiation aids in finding the slopes of the tangent lines at specified points.
Trigonometric Functions
Trigonometric functions are extensively employed in mathematics due to their recurring nature, especially in periodic phenomena. These functions, such as sine, cosine, and tangent, are based on ratios from the sides of right triangles. In our exercise, we focus on the function \( \tan x \), which is the ratio of the opposite side to the adjacent in a right triangle. Trigonometric functions are cyclic, repeating values over specific intervals, which is crucial for analyzing wave patterns and oscillations. The derivative of each of these baseline functions exhibits a unique trigonometric identity. For instance, the derivative of \( \tan x \) is \( \sec^2 x \), while the derivative of a composition like \( \tan 2x \) requires us to employ the chain rule, leading to the derivative \( 2 \sec^2 2x \). Understanding these derivatives helps in constructing precise models and evaluations for numerous real-world applications.
Derivative Evaluation
Derivative evaluation is the process of calculating the value of a derivative function at a specific point. This is essential for determining the exact slope of a tangent line at that point on the curve. In our exercise, we perform derivative evaluations to find slopes at given \( x \)-coordinates. For example, for \( f(x) = \tan x \) at \( x = \frac{\pi}{4} \), we substitute into \( f'(x) = \sec^2 x \) to obtain \( f'\left(\frac{\pi}{4}\right) = 2 \). This value represents how much the tangent line to \( f(x) \) rises or falls as \( x \) increases by one unit. Similarly, by evaluating the derivative of the function \( 6\tan x - \tan 2x \) at \( x = 0 \), we determine the slope to be 4. Evaluating the derivative accurately is vital for deriving the exact equation of a tangent line, assisting in predicting and understanding the behavior of various functions at specific points.

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