Find the derivative of the function y = 9 sec- (x + 5). (Use symbolic notation and fractions where needed.) y
Find the derivative of the function y = 9 sec- (x + 5). (Use symbolic notation and fractions where needed.) y
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Finding the Derivative of the Function
**Problem Statement:**
Find the derivative of the function \( y = 9 \sec^{-1} (x + 5) \).
**Instructions:**
Use symbolic notation and fractions where needed.
**Solution:**
\[
y' = \boxed{}
\]
In this context, \( \sec^{-1} \) denotes the inverse secant function. To find the derivative, apply the chain rule and the derivative formula for the inverse secant function:
The derivative of \( \sec^{-1}(x) \) is \( \frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{|x|\sqrt{x^2 - 1}} \).
For the function \( y = 9 \sec^{-1} (x + 5) \), we let \( u = x + 5 \), then the derivative of \( y \) with respect to \( x \) is given by:
\[
\frac{dy}{dx} = 9 \cdot \frac{d}{du}[\sec^{-1}(u)] \cdot \frac{du}{dx}
\]
Substitute \( u = x + 5 \):
\[
\frac{dy}{dx} = 9 \cdot \frac{1}{|x + 5| \sqrt{(x + 5)^2 - 1}} \cdot 1
\]
So the final expression for \( y' \) is:
\[
y' = \frac{9}{|x + 5| \sqrt{(x + 5)^2 - 1}}
\]
Therefore, the solution is:
\[
y' = \boxed{\frac{9}{|x+5| \sqrt{(x+5)^2 - 1}}}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7369c9e1-cd27-40fa-b19e-4e793313e403%2F41492ffe-4e33-4fe5-8ff9-810ceee929e6%2Fbhq7cal.png&w=3840&q=75)
Transcribed Image Text:### Finding the Derivative of the Function
**Problem Statement:**
Find the derivative of the function \( y = 9 \sec^{-1} (x + 5) \).
**Instructions:**
Use symbolic notation and fractions where needed.
**Solution:**
\[
y' = \boxed{}
\]
In this context, \( \sec^{-1} \) denotes the inverse secant function. To find the derivative, apply the chain rule and the derivative formula for the inverse secant function:
The derivative of \( \sec^{-1}(x) \) is \( \frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{|x|\sqrt{x^2 - 1}} \).
For the function \( y = 9 \sec^{-1} (x + 5) \), we let \( u = x + 5 \), then the derivative of \( y \) with respect to \( x \) is given by:
\[
\frac{dy}{dx} = 9 \cdot \frac{d}{du}[\sec^{-1}(u)] \cdot \frac{du}{dx}
\]
Substitute \( u = x + 5 \):
\[
\frac{dy}{dx} = 9 \cdot \frac{1}{|x + 5| \sqrt{(x + 5)^2 - 1}} \cdot 1
\]
So the final expression for \( y' \) is:
\[
y' = \frac{9}{|x + 5| \sqrt{(x + 5)^2 - 1}}
\]
Therefore, the solution is:
\[
y' = \boxed{\frac{9}{|x+5| \sqrt{(x+5)^2 - 1}}}
\]
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