Solving for f',f'',f''' and determining a general formula

In summary, to solve for f', f'', and f''', you will need to use the chain rule, product rule, and quotient rule in calculus. The general formula for f' can be determined by finding the derivative of the given function and simplifying it by factoring and combining like terms. You can use a calculator to find derivatives, but understanding the rules and process is important for accuracy. Finding derivatives can help understand the behavior of a function and can also be used to find critical points and slope of a tangent line. Tips for simplifying the general formula for f' include using algebraic rules and practicing with different types of functions.
  • #1
ttpp1124
110
4
Homework Statement
Can someone confirm if my work is correct?
Relevant Equations
n/a
IMG_4214.jpg
 
Physics news on Phys.org
  • #2
I'm a bit surprised you did not do what you were asked to do! The problem specifically asked you to "find [itex]f'(x)[/itex], [itex]f''(x)[/itex], [itex]f'''(x)[/itex], and [itex]f^{IV}(x)[/itex] then the nth derivative but you have only found [itex]f'(x)[/itex] and [itex]f''(x)[/itex] before jumping to the nth derivative.
Other than the fact that you do not have [itex]f'''(x)= 2^3 e^{2x}[/itex] and [itex]f^{IV}(x)= 2^4e^{2x}[/itex], every thing is good!
 
  • Like
Likes roam
  • #3
The problem also mentions "or ##f^{(n)}\left( \tfrac{1}{2}\right)##", which is of course ##\boxed{f^{(n)}\left( \tfrac{1}{2}\right) = 2^n e}##.
 

What is the purpose of solving for f', f'', and f'''?

The purpose of solving for f', f'', and f''' is to determine the first, second, and third derivatives of a given function f(x), respectively. These derivatives provide information about the rate of change, concavity, and inflection points of the function, which are useful in various applications such as optimization, curve sketching, and physics problems.

How do you solve for f'?

To solve for f', you can use the power rule, product rule, quotient rule, or chain rule, depending on the form of the function. The power rule states that the derivative of x^n is nx^(n-1). The product rule states that the derivative of f(x)g(x) is f'(x)g(x) + f(x)g'(x). The quotient rule states that the derivative of f(x)/g(x) is (f'(x)g(x) - f(x)g'(x))/g(x)^2. The chain rule states that the derivative of f(g(x)) is f'(g(x))g'(x).

What is the difference between f', f'', and f'''?

The difference between f', f'', and f''' is the number of times the function has been differentiated. f' is the first derivative, which gives the slope of the tangent line to the function at a given point. f'' is the second derivative, which gives information about the concavity of the function. f''' is the third derivative, which gives information about the rate of change of the concavity.

How do you determine a general formula for f'?

To determine a general formula for f', you can use the rules of differentiation mentioned earlier and apply them to the given function. For example, if f(x) = x^2, then f'(x) = 2x. However, for more complex functions, you may need to use multiple rules and algebraic manipulation to find the general formula for f'.

What are some common mistakes when solving for f', f'', and f'''?

Some common mistakes when solving for f', f'', and f''' include forgetting to use the chain rule, making algebraic errors, and not simplifying the final answer. It is also important to pay attention to the domain of the function and make sure the derivative is defined for all values in that domain. Additionally, when using the product or quotient rule, it is important to correctly identify which function is f(x) and which is g(x).

Similar threads

  • Calculus and Beyond Homework Help
Replies
9
Views
944
  • Calculus and Beyond Homework Help
Replies
32
Views
2K
  • Calculus and Beyond Homework Help
Replies
16
Views
974
  • Calculus and Beyond Homework Help
Replies
3
Views
524
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
549
  • Calculus and Beyond Homework Help
Replies
3
Views
608
  • Calculus and Beyond Homework Help
Replies
1
Views
346
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
Back
Top