(a) Find the differential [tex]$dy$[/tex].
[tex]
\begin{array}{c}
y=e^{x / 5} \\
d y=\square
\end{array}
[/tex]
(b) Evaluate [tex]$dy$[/tex] for the given values of [tex]$x$[/tex] and [tex]$dx$[/tex].
[tex]$d y=\quad \begin{array}{l}
\quad x=0, \quad d x=0.05
\end{array}$[/tex]
Answer (1)
Find the derivative of y = e x /5 with respect to x using the chain rule: d x d y = 5 1 e x /5 .
Determine the differential d y by multiplying the derivative by d x : d y = 5 1 e x /5 d x .
Evaluate d y at x = 0 and d x = 0.05 : d y = 5 1 e 0/5 ( 0.05 ) = 0.01 .
The value of d y for the given values is 0.01 .
Explanation
Problem Analysis We are given the function y = e x /5 and we need to find the differential d y and then evaluate it at x = 0 and d x = 0.05 .
Finding the Differential To find the differential d y , we first need to find the derivative of y with respect to x , which is d x d y . Using the chain rule, we have: d x d y = d x d e x /5 = e x /5 ⋅ d x d ( x /5 ) = e x /5 ⋅ 5 1 = 5 1 e x /5 Then, the differential d y is given by: d y = d x d y d x = 5 1 e x /5 d x
Evaluating the Differential Now, we need to evaluate d y at x = 0 and d x = 0.05 . Plugging in these values, we get: d y = 5 1 e 0/5 ( 0.05 ) = 5 1 e 0 ( 0.05 ) = 5 1 ( 1 ) ( 0.05 ) = 5 0.05 = 0.01 So, d y = 0.01 when x = 0 and d x = 0.05 .
Final Answer Therefore, the differential d y is 5 1 e x /5 d x , and its value at x = 0 and d x = 0.05 is 0.01 .
Examples
In physics, if you're analyzing the change in voltage ( V ) across a circuit component as you slightly adjust the current ( I ), you might have a relationship V = f ( I ) . The differential d V helps you estimate how much the voltage changes for a small change in current, d I . For instance, if V = e I /5 , you can find d V to predict voltage variations for tiny current adjustments, crucial for precise circuit tuning and stability analysis.
