Calculating the Derivative of w with Respect to t

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Homework Help Overview

The discussion revolves around calculating the derivative of a function \( w \) with respect to \( t \), where \( w \) is defined in terms of \( x \) and \( y \), which are themselves functions of \( t \). The participants are exploring different methods for differentiation, including the chain rule and simplification techniques.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers whether to use the chain rule or an alternative approach involving partial derivatives. Some participants suggest that using the chain rule could yield the same result but may be simpler in this context.

Discussion Status

The discussion is active, with participants providing insights on differentiation techniques and the simplification of functions before differentiation. There is an exchange of ideas regarding the application of logarithmic and exponential properties, indicating a productive exploration of the topic.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available or the methods allowed. There is an emphasis on understanding the relationships between the functions involved.

oswald
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Homework Statement


Finda dw/dt
w=3xy/x²-y²
x=t3
y=e2t

Homework Equations


w=(3t3e2t)/(t6-e4t)


The Attempt at a Solution


Well,is there anothe way to solve this, instead of dw/dt; like dw/dx * dw/dt + dw/dy * dy/dt ?
 
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Sure. You could just the chain rule as well. You should get the same answer both ways. Just substituting like you did looks to be a little easier.
 
Chain Rule Exponential and logarithmic

f(x) = ln [ e^ln(x+1) ]
f' = ?
 
Simplify it a bit first to make your differentiation easier. What is ln(e^whatever)?
 
f(x) = ln { e^[ln(x+1)] }

well, i have this answer, but i don't understand
ln [ e^ln(x+1) ] = ln(x+1)

f'(x) = 1/(e^ln(x+1)) * e^ln(x+1) * 1/(x+1) = 1/(x+1)
 
The ln and exp functions are inverses of one another, so for any real number u, ln(eu) = u. This means that you can simplify your function before taking its derivative. Then, what you end up differentiating is much simpler.
 

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