- #1
kim
- 1
- 0
Need to find the derivative of:
f(t) = square root of 3t+2 / 4t-3
please help.
f(t) = square root of 3t+2 / 4t-3
please help.
F(t) = square root of 3t+2 / 4t-3
- Thread starter kim
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- Root Square Square root
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Homework Help Overview
The discussion revolves around finding the derivative of the function f(t) = square root of (3t + 2) / (4t - 3). Participants are exploring different interpretations of the function's structure and the appropriate methods for differentiation.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Participants inquire about the correct interpretation of the function, debating whether it is the square root of the entire fraction or just the numerator. Various differentiation techniques are suggested, including the Quotient Rule, Product Rule, implicit differentiation, and logarithmic differentiation.
Discussion Status
The discussion is active, with multiple differentiation strategies being proposed. Participants are engaging in clarifying the function's format and exploring different mathematical approaches without reaching a consensus on the best method.
Contextual Notes
There is some ambiguity regarding the function's notation, which may affect the differentiation approach. Participants are also considering various differentiation techniques in the context of the problem.
- #2
Hurkyl
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What have you tried?
- #3
mathmike
- 204
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is this sqrt{3t+2 / 4t-3} or sqrt{3t+2} / 4t-3
- #4
bomba923
- 759
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Well, decide whether
!
[tex]\frac{d}{{dt}}\left[ {\frac{{f\left( t \right)}}{{g\left( t \right)}}} \right] = \frac{{f'\left( t \right)g\left( t \right) - f\left( t \right)g'\left( t \right)}}{{\left[ {g\left( t \right)} \right]^2 }}[/tex]
And the Quotient Rule will help you solve this problem~!
![tex]\frac{d}{{dt}}\left[ {\frac{{f\left( t \right)}}{{g\left( t \right)}}} \right] = \frac{{f'\left( t \right)g\left( t \right) - f\left( t \right)g'\left( t \right)}}{{\left[ {g\left( t \right)} \right]^2 }}[/tex]
Last edited:
- #5
Jameson
- 4,533
- 13
Or the product rule... :)
[tex]\sqrt{3t+2}*(4t-3)^{-1}[/tex]
[tex]\sqrt{3t+2}*(4t-3)^{-1}[/tex]
- #6
R_Y_A_N
- 9
- 0
or do implicit differentiation:
let y = sqrt(3t + 2)/(4t-3)
and then (4t-3)y = sqrt(3t+2)
let y = sqrt(3t + 2)/(4t-3)
and then (4t-3)y = sqrt(3t+2)
- #7
Jameson
- 4,533
- 13
Well, touche then. I'll think of something more clever later...
- #8
R_Y_A_N
- 9
- 0
while you are thinking of something else...here's another idea:
logarithmic differentiation
lny = (1/2)ln(3t+2) - ln(4t-1)
logarithmic differentiation
lny = (1/2)ln(3t+2) - ln(4t-1)
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