Estimate Ay using differentials. .1/5 y = x5 e3x-1 a = 1, dx = 0.1 (Give your answer to four decimal places.)
Estimate Ay using differentials. .1/5 y = x5 e3x-1 a = 1, dx = 0.1 (Give your answer to four decimal places.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![### Estimating Δy Using Differentials
To estimate the change in \( y \) (denoted as \( \Delta y \)) using differentials, consider the function provided:
\[ y = x^{1/5} e^{3x-1} \]
We are given:
\[ a = 1 \]
\[ dx = 0.1 \]
To find \( \Delta y \approx dy \), follow these steps:
1. **Compute the first derivative \( \frac{dy}{dx} \) of the function \( y = x^{1/5} e^{3x-1} \):**
\[ y = f(x) = x^{1/5} e^{3x-1} \]
Using the product rule and chain rule for differentiation:
\[ \frac{dy}{dx} = \frac{d}{dx} \left( x^{1/5} \cdot e^{3x-1} \right) \]
\[ \frac{dy}{dx} = \left( \frac{d}{dx} x^{1/5} \right) e^{3x-1} + x^{1/5} \left( \frac{d}{dx} e^{3x-1} \right) \]
Where \( \frac{d}{dx} x^{1/5} = \frac{1}{5} x^{-4/5} \) and \( \frac{d}{dx} e^{3x-1} = 3 e^{3x-1} \):
\[ \frac{dy}{dx} = \frac{1}{5} x^{-4/5} e^{3x-1} + x^{1/5} \cdot 3 e^{3x-1} \]
\[ \frac{dy}{dx} = e^{3x-1} \left(\frac{1}{5} x^{-4/5} + 3 x^{1/5} \right) \]
2. **Evaluate the derivative at \( a = 1 \):**
\[ \frac{dy}{dx} \bigg|_{x=1} = e^{3 \cdot 1 - 1} \left(\frac{1}{5} \cdot 1^{-4/5} + 3 \cdot 1^{1/5} \right)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7369c9e1-cd27-40fa-b19e-4e793313e403%2F37b2f0fb-0ee7-4f66-8610-7c2175c36dd9%2Fthlb5u.png&w=3840&q=75)
Transcribed Image Text:### Estimating Δy Using Differentials
To estimate the change in \( y \) (denoted as \( \Delta y \)) using differentials, consider the function provided:
\[ y = x^{1/5} e^{3x-1} \]
We are given:
\[ a = 1 \]
\[ dx = 0.1 \]
To find \( \Delta y \approx dy \), follow these steps:
1. **Compute the first derivative \( \frac{dy}{dx} \) of the function \( y = x^{1/5} e^{3x-1} \):**
\[ y = f(x) = x^{1/5} e^{3x-1} \]
Using the product rule and chain rule for differentiation:
\[ \frac{dy}{dx} = \frac{d}{dx} \left( x^{1/5} \cdot e^{3x-1} \right) \]
\[ \frac{dy}{dx} = \left( \frac{d}{dx} x^{1/5} \right) e^{3x-1} + x^{1/5} \left( \frac{d}{dx} e^{3x-1} \right) \]
Where \( \frac{d}{dx} x^{1/5} = \frac{1}{5} x^{-4/5} \) and \( \frac{d}{dx} e^{3x-1} = 3 e^{3x-1} \):
\[ \frac{dy}{dx} = \frac{1}{5} x^{-4/5} e^{3x-1} + x^{1/5} \cdot 3 e^{3x-1} \]
\[ \frac{dy}{dx} = e^{3x-1} \left(\frac{1}{5} x^{-4/5} + 3 x^{1/5} \right) \]
2. **Evaluate the derivative at \( a = 1 \):**
\[ \frac{dy}{dx} \bigg|_{x=1} = e^{3 \cdot 1 - 1} \left(\frac{1}{5} \cdot 1^{-4/5} + 3 \cdot 1^{1/5} \right)
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