5. Determine the equation of the tangent line for y = sech(In x) + tan='(x²) tangent at x = e*.A. Yr = 1.6071+ 6.5796 × 10-4(x – e4)C. yr = 1.6430x – 6.5796 x 10-4D. -yr = -1.6071 + 6.5796 x 10-4(x – e4)B. yT = 1.6430 + 6.5796 × 10-4x%3D

Name: 5. Determine the equation of the tangent line for y = sech(In x) + ta
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Description: 5. Determine the equation of the tangent line for y = sech(In x) + tan='(x²) tangent at x = e*.A. Yr = 1.6071+ 6.5796 × 10-4(x – e4)C. yr = 1.6430x – 6.5796 x 10-4D. -yr = -1.6071 + 6.5796 x 10-4(x – e4)B. yT = 1.6430 + 6.5796 × 10-4x%3D

Answered: Determine the equation of the tangent…

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter3: The Derivative

Section3.4: Definition Of The Derivative

Problem 6YT

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5. Determine the equation of the tangent line for y = sech(In x) + tan='(x²) tangent at x = e*.
A. Yr = 1.6071+ 6.5796 × 10-4(x – e4)
C. yr = 1.6430x – 6.5796 x 10-4
D. -yr = -1.6071 + 6.5796 x 10-4(x – e4)
B. yT = 1.6430 + 6.5796 × 10-4x
%3D

Expert Solution

Step 1

The equation of the tangent line for the function $f (x)$ at the point $x = a$ is, $y - f (a) = f' (a) (x - a)$ .

The function is $y = sech (\ln x) + \tan^{- 1} (x^{2})$ and the point is $x = e^{4}$ .

First find the value of $y$ at the point $x = e^{4}$ .

$\begin{array}{rcl} y & = & sech (\ln x) + \tan^{- 1} (x^{2}) \\ = & sech (\ln e^{4}) + \tan^{- 1} ({(e^{4})}^{2}) \\ = & sech (4) + \tan^{- 1} (e^{8}) \\ = & \tan^{- 1} (e^{8}) + \frac{2}{e^{- 4} + e^{4}} \\ \approx & 1.6071 \end{array}$

Step 2

Find the slope of the function by find the derivative of the function at the point $x = e^{4}$ .

$\begin{array}{rcl} y' & = & \frac{d}{d x} (sech (\ln (x)) + \tan^{- 1} (x^{2})) \\ = & \frac{d}{d x} (sech (\ln (x))) + \frac{d}{d x} (\tan^{- 1} (x^{2})) \\ = & - \frac{sech (\ln (x)) \tanh (\ln (x))}{x} + \frac{2 x}{x^{4} + 1} \\ y' (e^{4}) & = & - \frac{sech (\ln (e^{4})) \tanh (\ln (e^{4}))}{e^{4}} + \frac{2 e^{4}}{{(e^{4})}^{4} + 1} \\ \approx & 6.5796 \times 10^{- 4} \end{array}$

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Determine the equation of the tangent line for y = sech(In x) + tan-(x²) tangent at x = e*. A. Yr = 1.6071+6.5796 × 10–(x – e*) C. yT = 1.6430x – 6.5796 × 10-4 D. -yr = -1.6071+6.5796 × 10-(x – e*) B. yT = 1.6430 + 6.5796 × 10-4x

Determine the equation of the tangent line for y = sech(In x) + tan-(x²) tangent at x = e. A. Yr = 1.6071+6.5796 × 10–(x – e) C. yT = 1.6430x – 6.5796 × 10-4 D. -yr = -1.6071+6.5796 × 10-(x – e*) B. yT = 1.6430 + 6.5796 × 10-4x