Break down simple derivative problem for me

In summary, the conversation discusses the difficulty in understanding and applying the power rule for derivatives, specifically when dealing with cube roots. The speaker offers a brief explanation of the power rule and suggests practicing with problems to improve understanding and proficiency.
  • #1
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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  • #2
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.
 
  • #3
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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  • #4
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
 

Related to Break down simple derivative problem for me

1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. In simpler terms, it is the slope of a tangent line to a curve at a given point.

2. How do you break down a simple derivative problem?

To break down a simple derivative problem, you first need to identify the function and the variable that the derivative will be taken with respect to. Then you can use the power rule, product rule, quotient rule, or chain rule to find the derivative. Finally, plug in the given value for the variable to find the specific derivative at that point.

3. What is the power rule?

The power rule states that the derivative of a function raised to a constant power is equal to the constant power multiplied by the original function raised to the power minus one. In other words, if the function is f(x) = x^n, then its derivative is f'(x) = nx^(n-1).

4. How do you use the product rule to find a derivative?

The product rule states that the derivative of a product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function. In equation form, (fg)' = f'g + fg'.

5. Can you explain the chain rule?

The chain rule is used to find the derivative of a composite function, where one function is applied to the output of another function. It states that the derivative of a composite function f(g(x)) is equal to the derivative of the outer function f'(g(x)) multiplied by the derivative of the inner function g'(x). In equation form, (f(g(x)))' = f'(g(x)) * g'(x).

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