Solving 2 Chain Rule Problems: Struggling with Derivatives and Need Help

In summary, the conversation was about two difficult derivative problems using the chain rule. The first problem involved simplifying 9^(5-x^2) and the second problem involved finding the derivative of r/(r^2+5)^1/2. The equations used were d/dx(ax) = a^x ln a and d/dx(u/v) = (vdu/dx - udv/dx)/v^2. After some discussion and clarification, the summary of the solutions were given as -9^(5-x^2)2xln(9) and 5/(r^2+5)^3/2 respectively.
  • #1
bblair3
13
0
For some reason I am struggling with these problems. I am lost as a goose trying to fly south for the winter!

Homework Statement



9^(5-x2)
and
another derivative problem using chain rule
r/square root of the whole term r^2+5

Homework Equations



1st equation= d/dx= a^x ln a
2nd equation=

The Attempt at a Solution



1st one= -9^(5-x^2) 2x log (9)

2nd one= i am stuck at square root of (r^2 +5) + square rt (r^2+5) -r^2 all divided by sq root of (r^2+5) all divided by (r^2 +5)^2

any help would be greatly appreciated!
 
Physics news on Phys.org
  • #2
bblair3 said:
For some reason I am struggling with these problems. I am lost as a goose trying to fly south for the winter!

Homework Statement



9^(5-x^2)

and
another derivative problem using chain rule
r/square root of the whole term r^2+5

Homework Equations



1st equation= d/dx(ax) = a^x ln a
2nd equation=

The Attempt at a Solution



1st one= -9^(5-x^2) 2x log (9)

2nd one= i am stuck at square root of (r^2 +5) + square rt (r^2+5) -r^2 all divided by sq root of (r^2+5) all divided by (r^2 +5)^2

any help would be greatly appreciated!

1st one looks right.

2nd one: Write using a fractional exponent for square root.

r/(r2+5)1/2 = r(r2+5)-1/2. Take the derivative of this.
 
  • #3
apparently webassign does not like the answer -9^(5-x^2)2x log (9)

2. 5/(r^2+5)^3/2
 
  • #4
the second is right :) thanks...now to figure out what is going on with the first one
 
  • #5
Does WebAssign want log or does it want ln for the log base e ?
 
  • #6
i will try ln base e
 
  • #7
it indeed up being ln
thanks for your help Sammy S :)
 

1. What is a 2 chain rule problem?

A 2 chain rule problem is a type of mathematical problem that involves using the chain rule twice to find the derivative of a composite function. It is typically used when a function is composed of two or more functions inside of each other.

2. How do you solve a 2 chain rule problem?

To solve a 2 chain rule problem, you must first identify the inner and outer functions. Then, you can use the chain rule formula to find the derivatives of each function. Finally, you can combine the derivatives using the chain rule again to find the final derivative of the composite function.

3. When should I use the 2 chain rule?

The 2 chain rule is used when a function is composed of two or more functions inside of each other, and the chain rule needs to be applied more than once to find the derivative. It is commonly used in calculus and other areas of mathematics.

4. What are some common mistakes when solving 2 chain rule problems?

Some common mistakes when solving 2 chain rule problems include not correctly identifying the inner and outer functions, forgetting to use the chain rule formula, and making errors when combining derivatives using the chain rule. It is important to carefully follow the steps and double check your work to avoid these mistakes.

5. Are there any tips for solving 2 chain rule problems?

Some tips for solving 2 chain rule problems include practicing with various examples, clearly identifying the inner and outer functions, and using the chain rule formula correctly. It can also be helpful to break down the problem into smaller steps and double check your work at each step to catch any mistakes.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
813
  • Calculus and Beyond Homework Help
Replies
4
Views
910
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
Back
Top