differentiation of the following equation 3 i already done the differentiation, just wanna confirm either i got it right or wrong. Can someone verify this for me.
1) f(x) = $ -3\over x^{5/2}$
f '(x) 
= $ 3({ 5\over 2}x^{3/2})$ . $\frac{1}{x^5}$
= $ { 15\over 2}x^{(3/2)-5}$ 
= ${ 15\over 2}x^{-7/2}$ 
2) f(x) = $\frac{2x^2 + 3}{(x^3 - 4)^3}$
f ' (x) 
$= \frac{(4x) (x^3 - 4)^3  - (2x^2 + 3)[3(x^3 -4)^2 (3x)}{[(x^3 - 4)^3]^2}$
$= \frac{(4x) (x^3 - 4)^3  - 9x^2 (2x^2 + 3)(x^3 -4)^2}{(x^3 - 4)^6}$
$= \frac{(4x) (x^3 - 4)^3  - 9x^2 (2x^2 + 3)}{(x^3 - 4)^4}$  ---quotient rule
3) f(x) = sin(x cos x)
$f ' (x) = \cos [x \cos(x)] (\cos x - x \sin x)$
4) f(x) = $x^2$ tan 2x
$f ' (x) =  2x \tan (2x)$ + $2x^2$ $\sec^2 2x$
5) f(x) = $ 3 \ln (\cos 3x) $
$f ' (x) =  -9 \tan 3x$

1) $$\bigg( \frac{-3}{x^{5/2}} \bigg)'=-3x^\frac{-5}{2}=\frac{15}{2}
x^\frac{-7}{2}$$
2) $$ \bigg(\frac{2x^2+3}{(x^3-4)^3}\bigg)' =\bigg( (2x^2+3
)(x^3-4)^{-3}\bigg)' $$ $$= (4x)(x^3-4)^{-3} + (2x^2+3
)(-3)(x^3-4)^{-4}(3x^2) $$$$= (x^3-4)^{-3} [ 4x -9x^2(2x^2+3 )(x^3-4)^{-1}]
$$
