Derivative from Definition (square roots)

Click For Summary
SUMMARY

The discussion focuses on finding the derivative from the definition of the function f(x) = x + √x. The user correctly sets up the difference quotient as ((x+h) + √(x+h)) - (x + √x) / ((x+h) - x), which simplifies to (h + (√(x+h) - √x)) / h. The next step involves splitting this into two fractions and applying the limit as h approaches 0. The use of the conjugate method is suggested to simplify the square root term effectively.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with the definition of a derivative
  • Knowledge of algebraic manipulation, particularly with square roots
  • Experience with the conjugate multiplication technique
NEXT STEPS
  • Study the formal definition of a derivative in calculus
  • Learn about the limit process as h approaches 0
  • Explore the conjugate method for simplifying expressions involving square roots
  • Practice additional derivative problems using the difference quotient
USEFUL FOR

Students studying calculus, particularly those learning about derivatives, and educators seeking to clarify the derivative definition process.

I_LuV_FiZiX
Messages
27
Reaction score
0

Homework Statement


Find the derivative from definition of the functin f(x)= x + squareroot(x)


Homework Equations





The Attempt at a Solution


I only got as far as where I canceled out my positive and negative x terms in the numerator, and am left with three terms (2 of which are square roots, the other being an h), and I am wondering what to do now; is there some way to multiply by the conjugate?
 
Physics news on Phys.org
The difference quotient should look like

[tex] \frac{((x+h) + \sqrt{x+h}) - (x + \sqrt x)}{(x+h)-x} = \frac{h + (\sqrt{x+h} - \sqrt x)}{h},[/tex]

does it not? try splitting it into 2 fractions (based on the grouping I've shown above). One of the fractions is quite simple, the other you can simplify with the method you mentioned in your post. After the fractions are simplified, bring in the limits as [tex]h \to 0[/tex]
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
1K
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K