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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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:
$\displaystyle y = (x^2 + 2x)^5$
$\displaystyle \Rightarrow y' = 5(x^2 + 2x)^4 \cdot (2x + 2) = 10(x + 1)(x^2 + 2x)^4$
here, apply the product rule(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 $\displaystyle g(x) = 2x^3 - 5x^2 + 4x$
note that you can write the radicals as powers to obtain the following
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
two main options here.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