Can anyone show me how to defrentiate these 4 functions please:)

(3x^2-4x)^3

(x^2+2)(2x^3-5x^2+4x)

sqrt X - 3rd square root x not 3(sqrt x) but the 3 is very small and close to the square root sign

and 3x/x^2+4

*

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- Jan 31st 2009, 07:56 PMnotoriousmcCalculus Defrentiation
Can anyone show me how to defrentiate these 4 functions please:)

(3x^2-4x)^3

(x^2+2)(2x^3-5x^2+4x)

sqrt X - 3rd square root x not 3(sqrt x) but the 3 is very small and close to the square root sign

and 3x/x^2+4

* - Jan 31st 2009, 08:09 PMJhevon
hopefully you know the power rule for derivatives.

are you aware of the chain rule?

it says $\displaystyle \frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

here, your $\displaystyle f(x) = x^3$ and your $\displaystyle g(x) = 3x^2 - 4x$

an example:

$\displaystyle y = (x^2 + 2x)^5$

$\displaystyle \Rightarrow y' = 5(x^2 + 2x)^4 \cdot (2x + 2) = 10(x + 1)(x^2 + 2x)^4$

Quote:

(x^2+2)(2x^3-5x^2+4x)

it says (if $\displaystyle f$ and $\displaystyle g$ are functions of $\displaystyle x$), then

$\displaystyle \frac d{dx}[fg] = f'g + fg'$

here, your $\displaystyle f(x) = x^2 + 2$ and your $\displaystyle g(x) = 2x^3 - 5x^2 + 4x$

Quote:

sqrt X - 3rd square root x not 3(sqrt x) but the 3 is very small and close to the square root sign

$\displaystyle x^{1/2} - x^{1/3}$

now apply the power rule to find the derivative

Quote:

and 3x/x^2+4

(1) write as $\displaystyle 3x(x^2 + 4)^{-1}$ and use the product rule with $\displaystyle f(x) = 3x$ and $\displaystyle g(x) = (x^2 + 4)^{-1}$. note that you need the chain rule to find $\displaystyle g'(x)$

(2) use the quotient rule: $\displaystyle \frac d{dx} \left[ \frac fg \right] = \frac {f'g - fg'}{g^2}$

your $\displaystyle f$ and $\displaystyle g$ are the same as above - Jan 31st 2009, 08:11 PMGaloisTheory1
**Jhevon was faster than me.**

(3x^2-4x)^3= (3x^2-4x) (3x^2-4x) (3x^2-4x)

multiply it out and take the derivative

(x^2+2)(2x^3-5x^2+4x)

multiply it out and take the derivative

use quotient rule