Checking My Method For Differentiation

  • Thread starter BOAS
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  • #1
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!
 
Last edited:

Answers and Replies

  • #2
FeDeX_LaTeX
Gold Member
437
13
I want to use the chain rule here, so I would rearrange to this;

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

How did you get that?
 
  • #3
BOAS
555
19
[itex](x^{3})^{-4}[/itex] = [itex]4/x^{3}[/itex] is it not?
 
  • #4
FeDeX_LaTeX
Gold Member
437
13
[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}##
 
  • #5
36,311
8,281
[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.
 
  • #6
BOAS
555
19
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.
 
  • #7
BOAS
555
19
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]
 
  • #8
36,311
8,281
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.
 
  • #9
BOAS
555
19
Ah, I understand.

Thank you.
 

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