Checking My Method For Differentiation

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Homework Help Overview

The discussion revolves around differentiating the function y = 4/x³ + x³/4 with respect to x. Participants are exploring the application of differentiation rules, particularly the chain rule and power rule, in the context of this problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to rearrange the function for easier differentiation and are questioning the necessity of the chain rule. There is also a focus on clarifying the correct forms of the terms involved and their derivatives.

Discussion Status

There has been a productive exchange regarding the correct interpretation of the terms in the function. Some participants have pointed out mistakes in the original poster's reasoning, leading to a clearer understanding of the differentiation process. However, there is no explicit consensus on the final approach to the differentiation.

Contextual Notes

Participants are addressing potential misunderstandings regarding the application of the chain rule and the power rule, with some expressing concern over the treatment of constants in differentiation.

BOAS
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Homework Statement



Differentiate the following with respect to x

y = [itex]\frac{4}{x^{3}}[/itex] + [itex]\frac{x^{3}}{4}[/itex]

The Attempt at a Solution



So the problem here is really getting this into a form that is easy to differentiate and i'd just like to show what I'm doing before I go ahead and do the rest of my questions. The step I'm least confident with is applying chain rule to each term separately...

I want to use the chain rule here, so I would rearrange to this;

y = (x3)-4 + (4)-x3

dy/dx = -4(x3)-53x2 - x3(4)-x3-1.0

dy/dx = -4(x3)-53x2

dy/dx = -12x2(x3)-5

(made irrelevant by discussion below)

So, is this ok? I am concerned that my second term disappears, but that's what it looks like I have to do.

Thanks for any help you can give!
 
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BOAS said:
I want to use the chain rule here, so I would rearrange to this;

y = (x3)-4 + (4)-x3

How did you get that?
 
[itex](x^{3})^{-4}[/itex] = [itex]4/x^{3}[/itex] is it not?
 
BOAS said:
[itex](x^{3})^{-4}[/itex] = [itex]4/x^{3}[/itex] is it not?

It is not -- ##(x^3)^{-4} = x^{-12}##.

##\frac{4}{x^3} = 4x^{-3}##
 
BOAS said:
[itex](x^{3})^{-4}[/itex] = [itex]4/x^{3}[/itex] is it not?
No, it is not.

##\frac 4 {x^3} = 4x^{-3}##

For your problem there is no need for the chain rule.
 
FeDeX_LaTeX said:
It is not -- ##(x^3)^{-4} = x^{-12}##.

##\frac{4}{x^3} = 4x^{-3}##

I have made this mistake a few times before... Thanks for catching it.
 
Mark44 said:
No, it is not.

##\frac 4 {x^3} = 4x^{-3}##

For your problem there is no need for the chain rule.

That's become somewhat clearer to me after my mistake was pointed out.

so y = [itex]4x^{-3} + x^{3}(4^{-1})[/itex]

dy/dx = [itex]-12x^{-4} -4^{-2}3x^{2}[/itex]

dy/dx = [itex]-12x^{-4} - (3/2) x^{2}[/itex]
 
BOAS said:
That's become somewhat clearer to me after my mistake was pointed out.

so y = [itex]4x^{-3} + x^{3}(4^{-1})[/itex]

dy/dx = [itex]-12x^{-4} -4^{-2}3x^{2}[/itex]

dy/dx = [itex]-12x^{-4} - (3/2) x^{2}[/itex]

y = 4x-3 + (1/4)x3
dy/dx = -12x-4 + (3/4)x2

You are using the power rule on something that is not a function. 4-1 is a constant, so its derivative is zero.
 
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Ah, I understand.

Thank you.
 

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