Understanding the Derivative of F(x, y, z) = tanh (x+2y+3z)

  • Thread starter Thread starter makyol
  • Start date Start date
Click For Summary
The discussion focuses on finding the partial derivative of the function f(x, y, z) = tanh(x + 2y + 3z) with respect to x. The correct derivative is sech²(x + 2y + 3z), which is derived using the chain rule. The chain rule involves differentiating the outer function, tanh, and then multiplying it by the derivative of the inner function, u = x + 2y + 3z. A participant initially struggled with the concept but clarified their understanding after receiving guidance. The exchange highlights the importance of the chain rule in calculating derivatives of composite functions.
makyol
Messages
17
Reaction score
0
f(x, y, z) ?

Hey there,

Sorry for the title, but i cannot find any appropriate one for it, here is my question!

If f(x, y, z) = tanh (x+2y+3z) => fx =?(what is the derivative of f(x, y, z) with respect to x?)

Actually, i know the answer but i do not get the idea! In case it might help, the answer is that: sech^2(x+2y+3z)

So, as i said, can someone explain it to me? How can they reach this answer? Thanks!
 
Physics news on Phys.org


makyol said:
Hey there,

Sorry for the title, but i cannot find any appropriate one for it, here is my question!

If f(x, y, z) = tanh (x+2y+3z) => fx =?(what is the derivative of f(x, y, z) with respect to x?)

Actually, i know the answer but i do not get the idea! In case it might help, the answer is that: sech^2(x+2y+3z)

So, as i said, can someone explain it to me? How can they reach this answer? Thanks!
What you want is the partial derivative with respect to x. And you'll need the chain rule here, which looks something like this.
\frac{\partial tanh(u)}{\partial x} = \frac{\partial tanh(u)}{\partial u} \cdot \frac{\partial u}{\partial x}
 


Oh, i see now, i made a dummy mistake. Thank you for your quick reply!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
2
Views
2K
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
10
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K