Break down simple derivative problem for me

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Homework Help Overview

The discussion revolves around understanding derivatives, specifically the application of the power rule and its implications for various functions, including square roots and cube roots. The original poster expresses confusion about memorizing derivative formulas and seeks clarity on how to derive them, particularly for the cube root of a function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the reasoning behind derivative formulas and questions how to apply them to cube roots. Some participants mention the power rule and suggest deriving it from first principles, while others provide specific examples and transformations related to derivatives.

Discussion Status

Participants are exploring different interpretations of the power rule and its applications. Some guidance has been offered regarding the transformation of roots into exponent form, but there is no explicit consensus on the best approach to mastering these concepts.

Contextual Notes

The original poster indicates a reliance on memorization of derivative charts and expresses a desire for deeper understanding. There is mention of the need for practice and possibly seeking help from the teacher, highlighting the context of homework constraints.

member 5645
Our teacher simply tells us to memorize a chart of derivatives. This is all fine and dandy, until it came to our last test and I am lost.

I know [tex]d sqrt{X}=1/2X[/tex] and [tex]d 1/X^2=-2/X^3[/tex]
However, I don't know why that is, and it doesn't give me way to handle derivatives for cube roots, etc.
For example, what would the derivative of [tex]sqrt[3]{X^5}?[/tex]

Thanks for the help.


EDIT- apparently I'm not so adept at the Latex code. That last one should be cube root of X^5
 
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Ok those equations you describe are part of a much larger rule called the "power rule" (google for more info than I gave you)

basically,

a * (dx^p)/dx = a*p*x^(p-1) where a is a constant. dsqrt(x) is 1/2*x^(-1/2) however not 1/2X.

You can derive the power rule easily by using the definition of a derivative in limit form. It's related to the binomial expansion of (x+change in x)^p as change in x goes to 0. Anyway, you could find a better derivation online that I can provide since I can't use TeX to save my life.
 
OK, I think the best thing I can say here is,

[tex]\frac{d}{dx} x^n = nx^{(n-1)}[/tex]

Where n is any constant ( for a square root, n= 1/2, cube root, 1/3 etc.)

also, remember that

[tex]\sqrt[n]{x^m}=x^\frac{m}{n}[/tex]

This isn't from first principles, but it's a step in that direction from where you're at.

So, to answer your earlier question,

[tex]\frac{d}{dx}\sqrt[3]{x^5}[/tex]

[tex]=\frac{d}{dx} x^\frac{5}{3}[/tex]

[tex]= \frac{5}{3}x^{(\frac{5}{3}-1)}[/tex]

[tex]= \frac{5}{3}x^\frac{2}{3}[/tex]

[tex]= \frac{5}{3}\sqrt[3]{x^2}[/tex]
 
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My friend, this is one of those things where you can perfect yourself only by practice. The more problems you do and the less you try to look at the chart, the better you get at this part of calculus. Get hold of a set of problems + solutions (or better still, have your teacher help you out...he/she will be the best guide) and work them out. You should be adept at handling all this after some time.

Good Luck

Cheers
Vivek
 

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