How Do You Calculate a'(3) for a = f + 2g Given Derivative Values?

In summary: You're welcome. Let us see how far you get with the problem, and then we'll help you with further guidance.
  • #1
aizeltine
20
0

Homework Statement


Calculus Derivative help?
when x = 0, f=2, f '=1, g=5, g '= -4
when x = 1, f=3, f '=2, g=3, g '= -3
when x = 2, f=5, f '=3, g=1, g '= -2
when x = 3, f=10, f '=4, g=0, g '= -1
based on the above table, if a = f +2g, then a'(3) =?
please show steps.thanks


Homework Equations




if a = f +2g, then a'(3) =?

The Attempt at a Solution


a=f'+0(g')
a'(3)=f'3+0
a'(3)=4?
 
Physics news on Phys.org
  • #2
I don't understand what the problem is =(
 
  • #3
aizeltine said:
based on the above table, if a = f +2g, then a'(3) =?
Does the textbook give you the correct answer?
 
  • #4
aizeltine said:

Homework Statement


Calculus Derivative help?
when x = 0, f=2, f '=1, g=5, g '= -4
when x = 1, f=3, f '=2, g=3, g '= -3
when x = 2, f=5, f '=3, g=1, g '= -2
when x = 3, f=10, f '=4, g=0, g '= -1
based on the above table, if a = f +2g, then a'(3) =?
please show steps.thanks


Homework Equations




if a = f +2g, then a'(3) =?

The Attempt at a Solution


a=f'+0(g')
You didn't differentiate this correctly. You may want to review the properties of differentiation.
a'(3)=f'3+0
a'(3)=4?
 
  • #5

Homework Statement



Calculus Derivative help?
when x = 0, f=2, f '=1, g=5, g '= -4
when x = 1, f=3, f '=2, g=3, g '= -3
when x = 2, f=5, f '=3, g=1, g '= -2
when x = 3, f=10, f '=4, g=0, g '= -1
based on the above table, if a = f +2g, then a'(3) =?
please show steps.thanks


Homework Equations




if a = f +2g, then a'(3) =?

The Attempt at a Solution


a=f'+0(g')
a'(3)=f'3+0
a'(3)=4?


can anyone just please give me the steps on how to do it and just be direct
 
  • #6
Think about it.

a = f + 2g really means that a at 3 will equal f at 3 added to twice the value of g at 3. The latter information is given you in your table; just do the addition.

But how to calculate the derivative of a? That's where the point of the whole problem comes into play. The goal of the problem is to get you to think about one of the rules of derivatives. What rule have you learned that allows you to calculate the derivative of a sum of functions of which you already know the derivatives?
 
  • #7
aizeltine said:

Homework Statement



Calculus Derivative help?
when x = 0, f=2, f '=1, g=5, g '= -4
when x = 1, f=3, f '=2, g=3, g '= -3
when x = 2, f=5, f '=3, g=1, g '= -2
when x = 3, f=10, f '=4, g=0, g '= -1
based on the above table, if a = f +2g, then a'(3) =?
please show steps.thanks


Homework Equations

A very relevant equation is "(af(x)+ bg(x))'= af'(x)+ bg'(x)"


if a = f +2g, then a'(3) =?

The Attempt at a Solution


a=f'+0(g')
Where did this "0" come from? a'(3)= f'(3)+ 2g'(3)

a'(3)=f'3+0
a'(3)=4?


can anyone just please give me the steps on how to do it and just be direct
 
  • #8
aizeltine said:
can anyone just please give me the steps on how to do it and just be direct
That's not how it works here at Physics Forums. We provide help so that you can do a problem, but we don't do the work for you.
 
  • #9
Mark44 said:
That's not how it works here at Physics Forums. We provide help so that you can do a problem, but we don't do the work for you.
To the OP: also, stop posting the same question in multiple threads. I believe that's also against the forum rules.
 
  • #10
(Two threads merged)
 
  • #11
the eagle said:
well,i think this problem will be solved in the following steps:
Indeed it can be solved that way. But it may not help the OP's understanding of how to solve further exercises of this nature.

You do no one any favors by handing them the complete solution before they have had the opportunity to be guided to achieving it for themselves.
 
  • #12
you are right about that
thanks
 

What is a derivative in calculus?

A derivative in calculus represents the rate of change of a function at a specific point. It is a fundamental concept in calculus and is used to calculate the slope of a curve at a given point.

How do I find the derivative of a function?

To find the derivative of a function, you can use the rules of differentiation, such as the power rule, product rule, quotient rule, and chain rule. These rules allow you to calculate the derivative of a function by manipulating its algebraic form.

Why is the derivative important in calculus?

The derivative is important in calculus because it helps us understand how a function is changing at a particular point. It is used in many applications, such as optimization problems, physics, and economics, to determine the maximum or minimum values of a function.

What are some real-world applications of derivatives?

Some real-world applications of derivatives include determining the maximum profit for a business, calculating the velocity and acceleration of an object, and finding the optimum route for a vehicle to minimize travel time. It is also used in fields such as engineering, finance, and medicine.

What are the common mistakes when finding derivatives?

Some common mistakes when finding derivatives include forgetting to apply the chain rule, applying the power rule incorrectly, or forgetting to use the product or quotient rule when necessary. It is also important to carefully evaluate the limits and avoid algebraic errors when simplifying the expression.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
474
  • Calculus and Beyond Homework Help
Replies
6
Views
549
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
904
  • Calculus and Beyond Homework Help
Replies
4
Views
694
  • Calculus and Beyond Homework Help
Replies
2
Views
843
  • Calculus and Beyond Homework Help
Replies
3
Views
287
  • Calculus and Beyond Homework Help
Replies
2
Views
512
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
Back
Top