Solving a Derivative Problem with f(x) = SQRT(1-3x)

  • Thread starter Thread starter ladyrae
  • Start date Start date
  • Tags Tags
    Derivative
Click For Summary
SUMMARY

The derivative of the function f(x) = SQRT(1-3x) is calculated using the limit definition of the derivative. The process involves applying the limit as h approaches 0 to the expression [(f(x+h) - f(x))/h]. The final result is f'(x) = -3/(2f(x)). Understanding this fundamental concept is crucial for tackling more complex derivative problems in calculus.

PREREQUISITES
  • Understanding of the limit definition of a derivative
  • Familiarity with algebraic manipulation and simplification
  • Knowledge of square root functions and their properties
  • Basic calculus concepts, particularly derivatives
NEXT STEPS
  • Study the limit definition of derivatives in more depth
  • Practice derivative calculations for various functions
  • Explore the application of derivatives in real-world problems
  • Learn about higher-order derivatives and their significance
USEFUL FOR

Students learning calculus, educators teaching derivative concepts, and anyone seeking to strengthen their algebra and calculus skills.

ladyrae
Messages
32
Reaction score
0
My algebra continues to let me down…

How about this one…I know there a simpler ways to do these problems but at this point I’m supposed to do it the hard way.

Using the definition of derivate find f ` (x)

f(x) = SQRT(1-3x)

f ` (x) = lim h->0 [(f(x+h)) – (f(x))]/h

= lim->0 [(SQRT(1-3(x+h))) – (SQRT(1-3x))]/h

= lim->0 [[(SQRT(1-3(x+h))) – (SQRT(1-3x))]/h] . [[((SQRT(1-3x-3h))) + (SQRT (1-3x))]/ [((SQRT(1-3x-3h))) + (SQRT (1-3x))]]

=lim->0 (1-3x-3h-1+3x)/(h[(SQRT(1-3x-3h)) + (SQRT (1-3x))]

= lim->0 -3/[SQRT(1-3x-3h) + SQRT (1-3x)]

= -3/(2(SQRT(1-3x))
 
Physics news on Phys.org
Why are you let down by the correct answer? :smile:
 
)

= -3/(2f(x))

So, the derivative of f(x) is -3/(2f(x)).

I understand that this may seem like a difficult and tedious process, but it is important to understand the fundamentals of derivatives in order to solve more complex problems in the future. Keep practicing and don't get discouraged, algebra can be challenging but with persistence and effort, you will improve and become more confident in solving these types of problems. Good luck!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
24
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
8
Views
1K
Replies
34
Views
2K
Replies
40
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
12
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K