Master Calculus Questions with Expert Tips | Boost Your Grades

  • Context: MHB 
  • Thread starter Thread starter ADV1_
  • Start date Start date
  • Tags Tags
    Calculus Stuck
Click For Summary
SUMMARY

This discussion focuses on mastering calculus problems, specifically derivatives and tangent lines. For problem 8, the derivative of the function $f(x)=\sqrt{3x+25}$ is derived using the chain rule, resulting in $f'(x)=\frac{3}{2\sqrt{3x+25}}$. In problem 9, a correction is made regarding the evaluation of the function at $x=-1$, emphasizing the importance of notation in $f(-1)$. Problem 10 highlights the calculation of the slope of the tangent line at a specific point, reinforcing that it is determined by evaluating the derivative at that point.

PREREQUISITES
  • Understanding of derivatives and the power rule
  • Familiarity with the chain rule in calculus
  • Knowledge of evaluating functions at specific points
  • Ability to differentiate polynomial functions
NEXT STEPS
  • Study the application of the chain rule in more complex functions
  • Practice evaluating derivatives at specific points for various functions
  • Explore the concept of tangent lines and their slopes in calculus
  • Review common mistakes in function evaluation and notation
USEFUL FOR

Students studying calculus, educators teaching calculus concepts, and anyone looking to improve their understanding of derivatives and tangent lines.

ADV1_
Messages
1
Reaction score
0
unnamed(1).jpg
 

Attachments

  • unnamed2.jpg
    unnamed2.jpg
    168.7 KB · Views: 113
  • unnamed3.jpg
    unnamed3.jpg
    165.4 KB · Views: 104
Mathematics news on Phys.org
For problem 8, $f(x)=\sqrt{3x+25}$ which can be written as $f(x)=(3x+25)^{1/2}$. Now I suppose you know that the derivative of $x^n$ is $nx^{n- 1}$ (here n= 1/2) and that, by the chain rule, the derivative of $(g(x))^n= n(g(x))^{n-1} g'(x)$ (here g(x)= 3x+ 25 and g'(x)= 3).

For problem 9, yes, the slope of the the tangent line is 8. But you have the value of the function wrong! $f(-1)= -4(-1)^2$. That is NOT 4! (Look at the difference between $(-1)^2$ and $-1^2$.)

For problem 10 you have correctly found that the derivative is -2x+ 5 but you have left the next part, the slope of the tangent line at x=2, blank. Why is that? Do you not understand that the slope of the tangent line at a given x is the derivative at that x? Here that is -2(2)+5.
 

Similar threads

  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 45 ·
2
Replies
45
Views
8K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K