Solving the Derivative of f(a): A Frustrating Homework Problem

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Homework Help Overview

The discussion revolves around finding the derivative of the function f(t) = (2500 + 0.2t)(1 + t) / √(0.5t + 2). The original poster expresses frustration in obtaining the correct derivative, indicating a struggle with the application of the quotient rule.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the quotient rule and the chain rule, with some suggesting to simplify the numerator first. There is also a mention of confusion regarding variable notation, specifically between f(a) and f(t).

Discussion Status

Some participants have offered guidance on how to approach the problem, including expanding the numerator and taking derivatives separately. However, there is still uncertainty as one participant questions the correctness of their derived expression.

Contextual Notes

There is a noted confusion regarding variable notation, which may affect the clarity of the problem setup. Additionally, one participant mentions a lack of familiarity with LaTeX, which could impact the communication of mathematical expressions.

fghtffyrdmns
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Homework Statement


For some reason, I cannot seem to get the derivative for this.


Homework Equations



f(a) =\frac {(2500+0.2t)(1+t)}{\sqrt {0.5t+2}}

The Attempt at a Solution



f(a) =\frac {(2500+0.2t)(1+t)}{\sqrt {0.5t+2}}

c1fb89063c9c2285cffa29f34ca44490.png


\frac {(2500+0.2t)}{\sqrt {0.5t+2}} (1+t)

From here I would just use the quotient rule but I keep getting the wrong answer and I have no idea why.
 
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I'll simplify this for you. See if you can solve the question from here:

First, FOIL the numerator. Next, solve all of our components separately, so:

h'(x) = \frac{d}{dx} (0.5t + 2)1/2

= \frac{1}{2}(0.5t +2)-1/2 * 0.5 <By Chain Rule>

= \frac{1}{4} \frac{1}{\sqrt{(0.5t+2)}} <By x^-1 = 1/x>

g'(x) = 0.4t + 2500.2 <By power rule x^2 = 2x when the derivative is taken>

h(x)2 = 0.5t + 2 <x^1/2^2 = x^1>

next, plug in all of that data to the quotient rule, simplify, and see what you get!
 
ahhh, thank you!

at first, I thought I was supposed to treat it separately. Now, I just expanded the numerator and took the derivatives of both sides.

Thank you, sir :).
 
Both of you should get your variables straight.
fghtffyrdmns said:
f(a) =\frac {(2500+0.2t)(1+t)}{\sqrt {0.5t+2}}
That would be f(t), not f(a).
RPierre said:
h'(x) = \frac{d}{dx}(0.5t + 2)1/2
And that would be h&#039;(t) = \frac{d}{dt}(0.5t + 2)^{1/2}
 
Last edited:
I'm not familiar with LaTex and assumed he wasn't using leibniz notation by the prime notation used and the level of calculus, so I simply whipped up a solution with simple principles. I'm sure the original poster got the point as the question was able to be resolved.

Thanks for pointing that out though, I'll be more precise with my answers in the future, I'm also new to these forums.
 
ah yes, it's my fault. I wrote the wrong variables in accidentally.I got this as my answer

\frac {(0.5t+2)^{1/2}(0.4t+2500.2)-(2500+2500.2t+0.2t^{2})}{(2t+8)^{3/2}}
 
Last edited:
Hmm. This is not the right derivative. Where did I go wrong?
 

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