# Conjecture about the nth derivative of the function f(x)=e^(ax)

• Dustinsfl
In summary, the book is asking you to make a conjecture about the nth derivative of the function f(x) = e^(ax) that is based on patterns observed by playing around with all the derivatives and is self-contained, meaning it only involves the function and its derivatives. This conjecture should be written as a proposition with an appropriate quantifier.

#### Dustinsfl

Make a conjecture about the nth derivative of the function f(x)=e^(ax). This conjecture should be written as a self contained proposition including an appropriate quantifier.

What is the last sentence saying to do. I know what a conjecture is but I am confused on what the book wants here.

Dustinsfl said:
Make a conjecture about the nth derivative of the function f(x)=e^(ax). This conjecture should be written as a self contained proposition including an appropriate quantifier.

What is the last sentence saying to do. I know what a conjecture is but I am confused on what the book wants here.

To help you understand what "conjecture" means, consider the following small example. For example, let's pretend that I happen to notice that

$$1+2 = 3 = \frac{2 (2+1)}{2}$$

$$1+2+3 = 6 = \frac{3( 3+1)}{2}$$

$$1+2+3+4= 10 = \frac{4(4+1)}{2}$$

then I might hypothesize that

$$1+2+3+\dotsm + (n-1) + n = \frac{n(n+1)}{2}$$

Such an hypothesis (based on observed patterns) is a conjecture about the formula for the sum 1+2+3+...+(n-1)+n. It is an educated guess that appears to be true, but needs to be proved (or possibly disproved -- maybe our hypothesis is wrong).

So for your question, try playing around with all the derivatives (first, second, third, etc.) of the function e^(ax) and see if you can find any patterns. Then make a "conjecture" about the "nth derivative"

So the self contained proposition doesn't mean to do anything special or different then a just making any conjecture?

Dustinsfl said:
So the self contained proposition doesn't mean to do anything special or different then a just making any conjecture?

I believe that the intention of this problem is that the conjecture you make has only to do with the function f(x) = e^(ax) and its derivatives, and nothing more (self-contained). Otherwise, we could make up just about any conjecture.

## 1. What is the formula for the nth derivative of the function f(x)=e^(ax)?

The formula for the nth derivative of the function f(x)=e^(ax) is (a^n)e^(ax). This can also be written as a^n times the original function.

## 2. How do you find the nth derivative of a function?

To find the nth derivative of a function, you can use the power rule and product rule. First, take the nth derivative of the function's term with the highest power. Then, multiply that by the coefficient of that term. Next, take the (n-1)th derivative of the function's term with the second highest power, and so on. Finally, add all of these terms together to get the final formula for the nth derivative.

## 3. What role does the constant 'a' play in the nth derivative of f(x)=e^(ax)?

The constant 'a' determines the rate of change of the function f(x)=e^(ax). It acts as a scaling factor, meaning that a larger value of 'a' will result in a steeper slope and a faster rate of change.

## 4. Can the nth derivative of f(x)=e^(ax) be negative?

Yes, the nth derivative of f(x)=e^(ax) can be negative. This will depend on the value of 'a' and the power of 'n'. For example, if 'a' is negative and 'n' is an odd number, the nth derivative will be negative. However, if 'a' is negative and 'n' is an even number, the nth derivative will be positive.

## 5. Is there a limit to the number of times you can take the derivative of f(x)=e^(ax)?

Technically, there is no limit to the number of times you can take the derivative of f(x)=e^(ax). However, as the value of 'n' increases, the formula for the nth derivative becomes more complex and difficult to calculate. Furthermore, after a certain number of derivatives, the resulting function may become too complicated to be useful in practical applications.

• Calculus and Beyond Homework Help
Replies
36
Views
4K
• Calculus and Beyond Homework Help
Replies
16
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
899
• Calculus and Beyond Homework Help
Replies
8
Views
247
• Special and General Relativity
Replies
10
Views
511
• Calculus and Beyond Homework Help
Replies
6
Views
653
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
748
• General Math
Replies
1
Views
855
• Calculus and Beyond Homework Help
Replies
15
Views
1K