<s>[INST] 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) $
$f ' (x) =  -9 \tan 3x$
 [/INST] $$(-\frac3{x^{5/2}})'=-3x^{-5/2}=-3(-\frac52)x^{-\frac52-1}=\frac{15}{2x^{7/2}}$$
$$\left(\frac{2x^2+3}{(x^3-4)^3}\right)'=\frac{(4x)(x^3-4)^3-(2x^2+3)\big(3(3x^2)(x^3-4)^2\big)}{((x^3-4)^3)^2}=\frac{4x(x^3-4)-9(2x^2+3)x^2}{(x^3-4)^4}=-x\frac{14x^3+27x+16}{(x^3-4)^4}$$
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