Solving a Tricky Chain Rule Question with Confusing Variables

Click For Summary
SUMMARY

The discussion focuses on applying the chain rule to the function \( g(x,y) = f(2 \cdot \ln(x+y), x^4 - 3y^2) \) with specific derivatives \( f_{y}(0,-3) = -2 \) and \( f_{x}(0,-3) = 3 \). The user is confused about evaluating \( g_{y}(0,-3) \) due to the logarithmic function becoming undefined for negative inputs. The correct interpretation of the logarithmic transformation is clarified, emphasizing that \( 2\ln(x+y) \) can yield valid outputs when \( x+y \) is squared. The user is guided to utilize the chain rule effectively to find the correct derivative values.

PREREQUISITES
  • Understanding of the chain rule in calculus
  • Familiarity with partial derivatives
  • Knowledge of logarithmic functions and their properties
  • Basic algebraic manipulation skills
NEXT STEPS
  • Practice solving chain rule problems with multiple variables
  • Review the properties of logarithmic functions, especially in calculus
  • Explore examples of evaluating derivatives of composite functions
  • Learn about the implications of undefined values in logarithmic expressions
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable functions and the chain rule, as well as anyone seeking to clarify the application of derivatives in complex scenarios.

Yankel
Messages
390
Reaction score
0
Hello,

I have a tricky chain rule question, I think understanding it is more difficult than solving.

For the function z=f(x,y) it is given that:

f_{y}(0,-3)=-2

and

\[f_{x}(0,-3)=3\]

so for the function

\[g(x,y)=f(2\cdot ln(x+y),x^{4}-3y^{2})\]

choose the correct answer:

(1)
\[g_{y}(0,-3)=18\]

(2)
\[g_{y}(0,-3)=-6\]

(3)
\[g_{y}(0,1)=18\]

(4)
\[g_{y}(0,1)=-6\]

(5)
Non of the above answers

I am confused slightly. I thought to call

\[u=2\cdot ln(x+y)\]
and
\[v=x^{4}-3y^{2}\]

but when I put x=0 and y=-3, I can't get a value for u, since I get a negative value under the ln.

Can you assist clearing this up ? Thank you ! :confused:
 
Physics news on Phys.org
Their writing of the function is a bit sloppy. Remember $\displaystyle \begin{align*} 2\ln{(x+y)} = \ln{ \left[ (x + y)^2 \right] } \end{align*}$, which means you CAN have values of x and y such that $\displaystyle \begin{align*} x + y < 0 \end{align*}$, as squaring will make them positive.
 
I used the chain rule, and I called u= 2ln(x+y) and v=x^4-3y^2

I don't know how to proceed from here.
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
899
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K