# Math Help - Calculus Defrentiation

1. ## Calculus 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

*

2. Originally Posted by notoriousmc
Can anyone show me how to differentiate these 4 functions please

(3x^2-4x)^3
hopefully you know the power rule for derivatives.

are you aware of the chain rule?

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

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

an example:

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

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

(x^2+2)(2x^3-5x^2+4x)
here, apply the product rule

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

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

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

sqrt X - 3rd square root x not 3(sqrt x) but the 3 is very small and close to the square root sign
note that you can write the radicals as powers to obtain the following

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

now apply the power rule to find the derivative

and 3x/x^2+4
two main options here.

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

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

your $f$ and $g$ are the same as above

3. Jhevon was faster than me.

Originally Posted by notoriousmc

(3x^2-4x)^3

(3x^2-4x)^3= (3x^2-4x) (3x^2-4x) (3x^2-4x)
multiply it out and take the derivative

Originally Posted by notoriousmc
(x^2+2)(2x^3-5x^2+4x)
(x^2+2)(2x^3-5x^2+4x)
multiply it out and take the derivative

Originally Posted by notoriousmc
Do you mean 3x/(x^2+4)?
use quotient rule