Help with a derivative and solving for Critical points

In summary: That would give a simpler expression.In summary, the conversation is about a person struggling with simplifying and solving the derivative of the equation (x^2)/(sqrt(x+1)). After receiving help with simplifying the derivative using the quotient rule, they are reminded to use their algebra skills to further reduce the fraction. It is also suggested to use the product rule instead of the quotient rule for a simpler expression.
  • #1
LLofantaine
2
0
I have the equation

(x^2)/(sqrt(x+1))

This one has been stumping me, not sure how to reduce the derivative properly much less solve for 0.

I get it down to this using the quotient rule:

((x+1)^(1/2)*2x - x^2*1/2(x+1)^(-1/2)) / (x+1)

Just started learning derivative a few weeks ago and they're still stumping me. I'm really terrible at simplifying these down. Once it's simplified down how would I solve for ZERO to find the crit points?

Any help would be appreciated.
 
Physics news on Phys.org
  • #2
LLofantaine said:
I have the equation

(x^2)/(sqrt(x+1))

This one has been stumping me, not sure how to reduce the derivative properly much less solve for 0.

I get it down to this using the quotient rule:

((x+1)^(1/2)*2x - x^2*1/2(x+1)^(-1/2)) / (x+1)

Just started learning derivative a few weeks ago and they're still stumping me. I'm really terrible at simplifying these down. Once it's simplified down how would I solve for ZERO to find the crit points?

Any help would be appreciated.
Hello LLofantaine. Welcome to PF !

The derivative is correct. The rest is just algebra.

((x+1)^(1/2)*2x - x^2*1/2(x+1)^(-1/2)) / (x+1) → [itex]\displaystyle \frac{2x\sqrt{x+1}-\displaystyle \frac{x^2}{2\sqrt{x+1}}}{x+1}[/itex]
 
  • #3
Thanks, so that's what the derivative looks like, is there anyway to simplify that down further? Also, my algebra skills are obviously horrible so I can't remember the first thing about setting and solving that for 0
 
  • #4
LLofantaine said:
Thanks, so that's what the derivative looks like, is there anyway to simplify that down further? Also, my algebra skills are obviously horrible so I can't remember the first thing about setting and solving that for 0
To reduce the fraction, multiply the "main" numerator & denominator by √(x+1) . That should allow you to simplify the numerator greatly.

(Yes. When students say they're having trouble doing the calculus, it's often the algebra skills that are lacking.)
 
  • #5
The simplest way to handle this is NOT to use the quotient rule but, instead, write it as [tex]x^2(x+ 1)^{-1/2}[/tex] and use the product rule.
 

Related to Help with a derivative and solving for Critical points

1. What is a derivative?

A derivative is a mathematical concept used to describe the rate of change of a function at a specific point. It represents the slope of a tangent line at that point and can be used to find the instantaneous rate of change of a function.

2. 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 help you find the derivative of a function by manipulating its algebraic expression.

3. What are critical points?

Critical points are points on a function where the derivative is either zero or undefined. These points are important in finding the minimum and maximum values of a function and can also be used to determine the concavity of a function.

4. How do I solve for critical points?

To solve for critical points, you need to find the derivative of the function and set it equal to zero. Then, solve for the variable to find the x-values of the critical points. You can also use the second derivative test to determine if the critical points are maximum or minimum values.

5. Why are critical points important?

Critical points are important because they help us determine the behavior of a function. They can be used to find the minimum and maximum values of a function, as well as the points of inflection, where the concavity of the function changes. They also play a crucial role in optimization problems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
198
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
Replies
9
Views
756
  • Calculus and Beyond Homework Help
Replies
1
Views
224
Replies
7
Views
555
  • Calculus and Beyond Homework Help
Replies
25
Views
457
  • Calculus and Beyond Homework Help
Replies
22
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Back
Top