Simple Question about Power Rule

[mikexz]

First of all, can I have the y value in my final derivative answer?
I am having some trouble understand power rule compared to chain rule.

E.g.
Find derv. of f(x)=[(x^2)+3]^2
Rather than using chain rule, why can't I do this...

let a = [(x^2)+3]
f'(x)= a^2 [using power rule]
= 2a
= 2[(x^2)+3] [sub a= [(x^2)+3]


ONE more question:

Is there an easy to understand proof for why substitution fails.
f(x)=2x^2
f'(x) = 2 f'(x) x^2 [I originally thought that since the derivative of constants are =0, it would =0 but that didn't make any sense)

I really don't have a deep understanding of the power rule. I am really unsure, plz help (also if you have any really good calculus textbooks in mind plz suggest them)

thanks

 

[o_O]

1. It does work. Just you would have to use the chain rule to your result:
\(\displaystyle f(x) = a^{2}\)
\(\displaystyle f'(x) = 2a \cdot a' \quad \mbox{Chain rule}\)
\(\displaystyle f'(x) = 2(x^{2} + 3)} \cdot (2x)\)

So you can't avoid the chain rule on this one :wink:

2. The rule says:
\(\displaystyle \frac{d}{dx} c f(x) = c \frac{d}{dx} f(x)\)

What it's saying is that you can pull out the constant and take the derivative of the rest. Notice how the constant is outside the d/dx, indicating that you leave the constant as it is. Using your example:

\(\displaystyle \frac{d}{dx} 2x^{2} = 2 \frac{d}{dx} \left( x^{2} \right) = 2 \cdot \left( 2x \right) = 4x\)

Power rule:
\(\displaystyle \frac{d}{dx} x^{n} = nx^{n - 1}\)

Where n is some constant. Don't be tricked into using this rule when dealing with expressions such as:
\(\displaystyle \frac{d}{dx} a^{x}\) a isn't the variable we're differentiating with respect to and the exponent isn't constant.
\(\displaystyle \frac{d}{dx} x^{x}\) the exponent isn't constant.

 

[mikexz]

thank you