\begin{align*} u & = x^2 &&& v & = (3x^2 - 4)^5 \\ {du \over dx} & = 2x &&& {dv \over dx} & = 5(3x^2 - 4)^4 (6x) \phantom{000000} [\text{Chain rule}] \\ & &&& & = 30x (3x^2 - 4)^4 \end{align*} \begin{align*} {d \over dx} [x^2 (3x^2 - 4)^5 ] & = (x^2) [ 30x (3x^2 - 4)^4 ] + (3x^2 - 4)^5 (2x) \phantom{000000} [\text{Product rule}] \\ & = 30x^3 (3x^2 - 4)^4 + 2x (3x^2 - 4)^5 \\ & = 2x(3x^2 - 4)^4 [ 15x^2 + (3x^2 - 4) ] \phantom{00000000000} [\text{Factorise common factors}] \\ & = 2x(3x^2 - 4)^4 (18x^2 - 4) \\ & = 2x(3x^2 - 4)^4 (2) (9x^2 - 2) \\ & = 4x (3x^2 - 4)^4 (9x^2 - 2) \phantom{0} \text{ (Shown)} \end{align*}

\begin{align*} u & = e^{2x} &&& v & = \sqrt{3x^2 + 4} \\ {du \over dx} & = 2 e^{2x} &&& & = (3x^2 + 4)^{1 \over 2} \\ & &&& {dv \over dx} & = {1 \over 2} (3x^2 + 4)^{-{1 \over 2}} . (6x) \phantom{000000} [\text{Chain rule}] \\ & &&& & = 3x (3x^2 + 4)^{-{1 \over 2}} \end{align*} \begin{align*} {d \over dx} \left( e^{2x} \over (3x^2 + 4)^{1 \over 2} \right) & = { (3x^2 + 4)^{1 \over 2} (2 e^{2x} ) - (e^{2x}) [3x (3x + 4)^{-{1 \over 2}} ] \over [ (3x^2 + 4)^{1 \over 2} ]^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { 2 e^{2x} (3x^2 + 4)^{1 \over 2} - 3x e^{2x} (3x + 4)^{-{1 \over 2}} \over 3x^2 + 4) } \times { (3x + 4)^{1 \over 2} \over (3x + 4)^{1 \over 2} } \\ & = { 2 e^{2x} (3x^2 + 4)^1 - 3x e^{2x} (3x + 4)^0 \over (3x^2 + 4)^{3 \over 2} } \\ & = { 2 e^{2x} (3x^2 + 4) - 3x e^{2x} (1) \over \sqrt{ (3x^2 + 4)^3 } } \\ & = { 6 x^2 e^{2x} + 8 e^{2x} - 3x e^{2x} \over \sqrt{ (3x^2 + 4)^3 } } \\ & = { e^{2x} ( 6x^2 - 3x + 8) \over \sqrt{ (3x^2 + 4)^3 } } \\ \\ \therefore a & = 6, b = -3, c = 8 \end{align*}

\begin{align*} {d \over dx} [ \tan^3 (4x) ] & = {d \over dx} [ \tan (4x) ] ^3 \\ & = 3[ \tan (4x) ]^2 (4 \sec^2 4x) \phantom{000000} [\text{Chain rule}] \\ & = 12 \tan^2 (4x) \sec^2 (4x) \end{align*}

\begin{align*} {2x - 1 \over (x - 1)^2} & = {A \over (x - 1)} + {B \over (x - 1)^2} \\ & = {A(x - 1) \over (x - 1)^2} + {B \over (x - 1)^2} \\ & = {A(x - 1) + B \over (x - 1)^2} \\ \\ 2x - 1 & = A(x - 1) + B \\ \\ \text{Let } & x = 1, \\ 2(1) - 1 & = A(0) + B \\ 1 & = B \\ \\ 2x - 1 & = A(x - 1) + 1 \\ \\ \text{Let } & x = 0, \\ 2(0) - 1 & = A(0 - 1) + 1 \\ -1 & = A(-1) + 1 \\ -1 & = -A + 1 \\ A & = 1 + 1 \\ & = 2 \\ \\ \therefore {2x - 1 \over (x - 1)^2} & = {2 \over (x - 1)} + {1 \over (x - 1)^2} \\ \\ \therefore A & = 2, B = 1 \end{align*}

\begin{align*} {d \over dx}\left[ {2x - 1 \over (x - 1)^2} \right] & = {d \over dx}\left[ {2 \over x - 1} + {1 \over (x - 1)^2} \right] \\ & = {d \over dx}[ 2(x - 1)^{-1} + (x - 1)^{-2} ] \\ & = {d \over dx}(2(x - 1)^{-1}) + {d \over dx}(x - 1)^{-2} \\ & = (2)(-1)(x - 1)^{-2}(1) + (-2)(x - 1)^{-3}(1) \\ & = -2(x - 1)^{-2} - 2(x - 1)^{-3} \\ & = -2 \left[ 1 \over (x - 1)^2 \right] - 2 \left[ 1 \over (x - 1)^3 \right] \\ & = {-2 \over (x - 1)^{2}} - {2 \over (x - 1)^3} \\ & = {-2(x - 1) \over (x - 1)^{3}} - {2 \over (x - 1)^3} \\ & = {-2x + 2 - 2 \over (x - 1)^{3}} \\ & = {-2x \over (x - 1)^{3}} \\ & = -{2x \over (x - 1)^3} \end{align*}

\begin{align*} y & = \ln (\sec x + \tan x) \\ \\ {d \over dx}(\sec x) & = {d \over dx} \left(1 \over \cos x\right) \\ & = {d \over dx} (\cos x)^{-1} \\ & = (-1)(\cos x)^{-2} (-\sin x) \\ & = \left(1 \over \cos x\right)^2 (\sin x) \\ & = {1 \over \cos x} \left(\sin x \over \cos x\right) \\ & = \sec x \tan x \\ \\ {dy \over dx} & = {\sec x \tan x + \sec^2 x \over \sec x + tan x} \\ & = {\sec x( \tan x + \sec x) \over \sec x + \tan x} \\ & = \sec x \\ \\ {d^2 y \over dx^2} & = \sec x \tan x \\ \\ \\ {d^2 y \over dx^2} - \tan x \left(dy \over dx\right) & = \sec x \tan x - \tan x (\sec x) \\ & = \sec x \tan x - \sec x \tan x \\ & = 0 \phantom{0} \text{ (Shown)} \end{align*}