dw For the given functions, (a) express as a function of t, both by using the chain rule and by expressing w in terms of t and dt dw differentiating directly with respect to t. Then (b) evaluate dt at the given value of t. w=7ye* - Inz, x= In (1²+1), y=tan¯¹t, z=e¹;t=1
dw For the given functions, (a) express as a function of t, both by using the chain rule and by expressing w in terms of t and dt dw differentiating directly with respect to t. Then (b) evaluate dt at the given value of t. w=7ye* - Inz, x= In (1²+1), y=tan¯¹t, z=e¹;t=1
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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![**Topic: Calculus - Differentiation Using the Chain Rule**
For the given functions,
1. **Express \(\frac{dw}{dt}\) as a function of \(t\)**, both by:
- Using the chain rule.
- Expressing \(w\) in terms of \(t\) and differentiating directly with respect to \(t\).
2. **Evaluate \(\frac{dw}{dt}\) at the given value of \(t\)**.
Given:
\[ w = 7y e^x - \ln z \]
\[ x = \ln \left( t^2 + 1 \right) \]
\[ y = \tan^{-1} t \]
\[ z = e^{t^2} \]
\[ t = 1 \]
**Step-by-Step Solution:**
1. **Using the Chain Rule:**
To find \( \frac{dw}{dt} \) using the chain rule, we express \( \frac{dw}{dt} \) in terms of partial derivatives with respect to \( x \), \( y \), and \( z \):
\[
\frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} + \frac{\partial w}{\partial z} \frac{dz}{dt}
\]
Given the functions:
\[
w = 7y e^x - \ln z
\]
2. **Express \(w\) in terms of \(t\) and differentiate directly:**
- \( x = \ln (t^2 + 1) \)
- \( y = \tan^{-1} t \)
- \( z = e^{t^2} \)
Substitute these into the equation for \( w \):
\[
w = 7 (\tan^{-1} t) e^{\ln (t^2 + 1)} - \ln (e^{t^2})
\]
Since \( e^{\ln (t^2 + 1)} = t^2 + 1 \) and \( \ln (e^{t^2}) = t^2 \):
\[
w = 7 (\tan^{-1} t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2d576798-e517-451f-b302-9c2c35ed41b2%2F59f8f6d9-0bbf-414b-90f1-5c71a25e16f6%2Fy44yorp_processed.png&w=3840&q=75)
Transcribed Image Text:**Topic: Calculus - Differentiation Using the Chain Rule**
For the given functions,
1. **Express \(\frac{dw}{dt}\) as a function of \(t\)**, both by:
- Using the chain rule.
- Expressing \(w\) in terms of \(t\) and differentiating directly with respect to \(t\).
2. **Evaluate \(\frac{dw}{dt}\) at the given value of \(t\)**.
Given:
\[ w = 7y e^x - \ln z \]
\[ x = \ln \left( t^2 + 1 \right) \]
\[ y = \tan^{-1} t \]
\[ z = e^{t^2} \]
\[ t = 1 \]
**Step-by-Step Solution:**
1. **Using the Chain Rule:**
To find \( \frac{dw}{dt} \) using the chain rule, we express \( \frac{dw}{dt} \) in terms of partial derivatives with respect to \( x \), \( y \), and \( z \):
\[
\frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} + \frac{\partial w}{\partial z} \frac{dz}{dt}
\]
Given the functions:
\[
w = 7y e^x - \ln z
\]
2. **Express \(w\) in terms of \(t\) and differentiate directly:**
- \( x = \ln (t^2 + 1) \)
- \( y = \tan^{-1} t \)
- \( z = e^{t^2} \)
Substitute these into the equation for \( w \):
\[
w = 7 (\tan^{-1} t) e^{\ln (t^2 + 1)} - \ln (e^{t^2})
\]
Since \( e^{\ln (t^2 + 1)} = t^2 + 1 \) and \( \ln (e^{t^2}) = t^2 \):
\[
w = 7 (\tan^{-1} t
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