Break down simple derivative problem for me

[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 d sqrt{X}=1/2X and d 1/X^2=-2/X^3
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 sqrt[3]{X^5}?

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

 

[vsage]

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.

 

[tyco05]

OK, I think the best thing I can say here is,

\frac{d}{dx} x^n = nx^{(n-1)}

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

also, remember that

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

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,

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

=\frac{d}{dx} x^\frac{5}{3}

= \frac{5}{3}x^{(\frac{5}{3}-1)}

= \frac{5}{3}x^\frac{2}{3}

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

 

[maverick280857]

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