Need help understanding a differentiation

In summary, there is a confusion with a step involving differentiation in a problem solution, and the speaker is unsure of where a certain group of terms in the solution comes from. There is also a typo in the solution. The second group of terms after the + sign is present because dx/ds is typically a function of t, not a constant. The product rule is needed when differentiating the result of the top half.
  • #1
Nat3
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I'm working my way through a solution of a problem and am confused on a step where a differentiation is performed. I'm sure I'm just forgetting some kind of rule, but I've been perusing my textbook and can't seem to figure out what I'm missing. Here's the step I'm talking about:

5nnt.png


Note that there's a typo in the solution: [itex]\frac{dx^2}{dt^2}[/itex] should be [itex]\frac{d^2x}{dt^2}[/itex].

The top part I understand completely, it's the bottom half I'm confused about. It seems to me that:

[itex]\frac{d^2x}{dt^2} = \frac{d}{dt}(\frac{dx}{dt}) = \frac{dx}{ds}(\frac{\alpha}{2}\sqrt{\frac{K}{m}}e^{-\frac{\alpha t}{2}})[/itex]

In other words, I don't understand where the second group of terms (after the + sign) comes from in the last line. Why is it there?
 
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  • #2
Nat3 said:
I'm working my way through a solution of a problem and am confused on a step where a differentiation is performed. I'm sure I'm just forgetting some kind of rule, but I've been perusing my textbook and can't seem to figure out what I'm missing. Here's the step I'm talking about:

5nnt.png


Note that there's a typo in the solution: [itex]\frac{dx^2}{dt^2}[/itex] should be [itex]\frac{d^2x}{dt^2}[/itex].

The top part I understand completely, it's the bottom half I'm confused about. It seems to me that:

[itex]\frac{d^2x}{dt^2} = \frac{d}{dt}(\frac{dx}{dt}) = \frac{dx}{ds}(\frac{\alpha}{2}\sqrt{\frac{K}{m}}e^{-\frac{\alpha t}{2}})[/itex]

In other words, I don't understand where the second group of terms (after the + sign) comes from in the last line. Why is it there?

It's because dx/ds is generally a function of t. It's not a constant. You need to use the product rule when you differentiate the result of the top half.
 
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Related to Need help understanding a differentiation

1. What is differentiation?

Differentiation is a mathematical concept that is used to find the rate at which one variable changes with respect to another variable. It is also known as the derivative and is an important tool in calculus.

2. Why is differentiation important?

Differentiation is important because it allows us to understand how a function is changing at a specific point. This information is crucial in many scientific fields, such as physics, engineering, and economics.

3. How is differentiation calculated?

Differentiation is calculated using a set of rules, known as differentiation rules, which involve taking the derivative of a function with respect to a variable. These rules include the power rule, product rule, quotient rule, and chain rule.

4. What is the difference between differentiation and integration?

Differentiation and integration are inverse operations of each other. While differentiation involves finding the rate of change of a function, integration involves finding the area under a curve. In other words, differentiation is the process of finding a derivative, while integration is the process of finding the anti-derivative.

5. How can differentiation be applied in real-world situations?

Differentiation can be applied in many real-world situations, such as calculating the speed and acceleration of an object, optimizing functions in engineering and economics, and predicting the growth of populations in biology. It is also used in data analysis to find patterns and trends in data sets.

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