(i)

$$\begin{align} y & = \ln (x^2 - 4) \\ \\ {dy \over dx} & = {2x \over x^2 - 4} \end{align}$$

(ii)

$$\begin{align} {dx \over dt} & = p \text{ units per second} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = {2x \over x^2 - 4} \times p \\ & = {2x \over x^2 - 4} \times {p \over 1} \\ & = {2p x \over x^2 - 4} \\ \\ \text{When } & x = 3, \\ {dy \over dt} & = {2p (3) \over (3)^2 - 4} \\ & = {6p \over 5} \\ & = {6 \over 5}p \text{ units per second} \end{align}$$

$$\begin{align} {dx \over dt} & = 3 \text{ units per second} \\ \\ y & = \ln (2 + e^{-x} ) \\ \\ {dy \over dx} & = { (-1)e^{-x} \over 2 + e^{-x} } \\ & = { - e^{-x} \over 2 + e^{-x} } \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = { - e^{-x} \over 2 + e^{-x} } \times 3 \\ & = { - e^{-x} \over 2 + e^{-x} } \times {3 \over 1} \\ & = { - 3 e^{-x} \over 2 + e^{-x} } \\ \\ \text{When } & x = 1, \\ {dy \over dt} & = { - 3e^{-1} \over 2 + e^{-1} } \\ & = { - 3e^{-1} \over 2 + e^{-1} } \times {e \over e} \\ & = { - 3 e^{-1 + 1} \over 2e + e^{-1 + 1} } \\ & = { - 3 e^0 \over 2e + e^0 } \\ & = { - 3(1) \over 2e + 1 }\\ & = {- 3 \over 2e + 1} \text{ units per second} \end{align}$$

(i)

$$\begin{align} u & = 2x &&& v & = \ln x \\ {du \over dx} & = 2 &&& {dv \over dx} & = {1 \over x} \end{align}$$ $$\begin{align} {dy \over dx} & = (2x)\left(1 \over x\right) + (\ln x)(2) \phantom{000000} [\text{Product rule}] \\ & = {2x \over 1}\left(1 \over x\right) + 2 \ln x \\ & = {2x \over x} + 2 \ln x \\ & = 2 + 2 \ln x \\ & = 2(1 + \ln x) \end{align}$$

(ii)

$$\begin{align} {dy \over dx} & = 2(1 + \ln x) \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 2(1 + \ln x) \\ {0 \over 2} & = 1 + \ln x \\ 0 & = 1 + \ln x \\ - \ln x & = 1 \\ \ln x & = -1 \\ \log_e x & = -1 \\ x & = e^{-1} \\ & = {1 \over e} \end{align}$$

(iii)

$$\begin{align} {dy \over dx} & = 2(1 + \ln x) \\ & = 2 + 2\ln x \\ \\ {d^2 y \over dx^2} & = 2 \left(1 \over x\right) \\ & = {2 \over 1} \left(1 \over x\right) \\ & = {2 \over x} \\ \\ \text{When } & x = {1 \over e}, \\ {d^2 y \over dx^2} & = {2 \over {1 \over e}} \\ & = 2 \div {1 \over e} \\ & = 2 \times e \\ & = 2e \end{align}$$

(i)

$$\begin{align} u & = \ln x &&& v & = x^2 \\ {du \over dx} & = {1 \over x} &&& {dv \over dx} & = 2x \end{align}$$ $$\begin{align} {dy \over dx} & = {(x^2) \left(1 \over x\right) - (\ln x)(2x) \over (x^2)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = {x - 2x \ln x \over x^4} \\ & = { x (1 - 2 \ln x) \over x^4 } \\ & = { 1 - 2\ln x \over x^3 } \end{align}$$

(ii)

$$\begin{align} {dy \over dx} & = { 1 - 2\ln x \over x^3 } \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {1 - 2 \ln x \over x^3} \\ 0 & = 1 - 2 \ln x \\ 2 \ln x & = 1 \\ \ln x & = {1 \over 2} \\ \log_e x & = {1 \over 2} \\ x & = e^{1 \over 2} \\ & = \sqrt{e} \end{align}$$

(iii)

$$\begin{align} {dy \over dx} & = { 1 - 2\ln x \over x^3 } \end{align}$$
$$\begin{align} u & = 1 - 2 \ln x &&& v & = x^3 \\ {du \over dx} & = - 2 \left(1 \over x\right) &&& {dv \over dx} & = 3x^2 \\ & = -{2 \over x} \end{align}$$ $$\begin{align} {d^2 y \over dx^2} & = { (x^3) \left(-{2 \over x}\right) - (1 - 2 \ln x)(3x^2) \over (x^3)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { {-2x^3 \over x} - 3x^2 (1 - 2 \ln x) \over x^6 } \\ & = { - 2x^2 - 3x^2 (1 - 2 \ln x) \over x^6 } \\ & = { x^2 [ - 2 - 3(1 - 2 \ln x) ] \over x^6 } \\ & = { - 2 - 3(1 - 2 \ln x) \over x^4 } \\ & = { - 2 - 3 + 6 \ln x \over x^4 } \\ & = { - 5 + 6 \ln x \over x^4 } \\ \\ \text{When } & x = e^{1 \over 2}, \\ {d^2 y \over dx^2} & = { - 5 + 6 \ln e^{1 \over 2} \over (e^{1 \over 2})^4 } \\ & = { - 5 + 6 \left(1 \over 2\right) \over e^2 } \\ & = { - 2 \over e^2} \end{align}$$

$$\begin{align} V & = \lambda x^2 \ln \left(1 \over x\right) \\ & = \lambda x^2 \ln x^{-1} \\ & = \lambda x^2 (-1) \ln x \phantom{00000000} [\text{Power law (logarithms)}] \\ & = -\lambda x^2 \ln x \end{align}$$
$$\begin{align} u & = - \lambda x^2 &&& v & = \ln x \\ {du \over dx} & = - \lambda (2)x &&& {dv \over dx} & = {1 \over x} \\ & = - 2\lambda x \end{align}$$ $$\begin{align} {dV \over dx} & = (-\lambda x^2)\left(1 \over x\right) + (\ln x)(- 2 \lambda x) \phantom{000000} [\text{Product rule}] \\ & = { -\lambda x^2 \over x } - 2 \lambda x \ln x \\ & = -\lambda x - 2\lambda x \ln x \end{align}$$
$$\begin{align} u & = - 2 \lambda x &&& v & = \ln x \\ {du \over dx} & = -2\lambda (1) &&& {dv \over dx} & = {1 \over x} \\ & = - 2 \lambda \end{align}$$ $$\begin{align} {d^2 V \over dx^2} & = - \lambda(1) + (-2 \lambda x)\left(1 \over x\right) + (\ln x)(-2 \lambda) \\ & = - \lambda + {-2\lambda x \over x} - 2 \lambda \ln x \\ & = - \lambda - 2 \lambda - 2 \lambda \ln x \\ & = - 3 \lambda - 2 \lambda \ln x \end{align}$$
$$\begin{align} \text{Let } & {dV \over dx} = 0, \\ 0 & = - \lambda x - 2\lambda x \ln x \\ 0 & = - \lambda x (1 + 2 \ln x) \end{align}$$ $$\begin{align} - \lambda x & = 0 && \text{ or } & 1 + 2 \ln x & = 0 \\ x & = 0 \text{ (Reject, since } 0 < x < 1) &&& 2 \ln x & = -1 \\ & &&& \ln x & = -{1 \over 2} \\ & &&& \log_e x & = -{1 \over 2} \\ & &&& x & = e^{-{1 \over 2}} \\ & &&& & = {1 \over \sqrt{e}} \\ \\ & &&& \text{Substitute } & \text{into } {d^2 V \over dx^2}, \\ & &&& {d^2 V \over dx^2} & = - 3 \lambda - 2\lambda \ln e^{-{1 \over 2}} \\ & &&& & = -3 \lambda - 2 \lambda \left(-{1 \over 2}\right) \\ & &&& & = -3 \lambda + \lambda \\ & &&& & = -2 \lambda \end{align}$$ $$\begin{align} \text{Since } \lambda > 0, & \phantom{.} {d^2 V \over dx^2} > 0 \\ \\ \therefore \text{When } x = {1 \over \sqrt{e}}, & \phantom{.} \text{ speed (} V)\text{ is a maximum} \end{align}$$

Note: I don't think this will be tested in O levels (more for A levels)

$$\begin{align} y & = 2^x \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln y & = \ln 2^x \\ \ln y & = x \ln 2 \phantom{000000} [\text{Power law (logarithms)}] \\ \ln y & = (\ln 2) x \\ \\ \text{Differentiate } & \text{both sides w.r.t } x, \\ {d \over dx} ( \ln y) & = {d \over dx} [ (\ln 2) x ] \\ {1 \over y} . {dy \over dx} & = (\ln 2)(1) \\ {1 \over y} {dy \over dx} & = \ln 2 \\ {dy \over dx} & = (\ln 2) y \phantom{00} \text{ (Shown)} \end{align}$$

Note: I don't think this will be tested in O levels (more for A levels)

$$\begin{align} y & = (2 + x^2)^3 (1 - x^3)^4 \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln y & = \ln \left[ (2 + x^2)^3 (1 - x^3)^4 \right] \\ \ln y & = \ln (2 + x^2)^3 + \ln (1 - x^3)^4 \phantom{000000} [\text{Product law (logarithms)}] \\ \ln y & = 3\ln (2 + x^2) + 4 \ln (1 - x^3) \phantom{00000} [\text{Power law (logarithms)}] \\ \\ \text{Differentiate } & \text{both sides w.r.t } x, \\ {d \over dx} ( \ln y) & = {d \over dx} \left[ 3\ln (2 + x^2) + 4 \ln (1 - x^3) \right] \\ {1 \over y} . {dy \over dx} & = 3 \left(2x \over 2 + x^2\right) + 4 \left(- 3x^2 \over 1 - x^3\right) \\ {1 \over y} . {dy \over dx} & = {6x \over 2 + x^2} - {12x^2 \over 1 - x^3} \\ {dy \over dx} & = y \left( {6x \over 2 + x^2} - {12x^2 \over 1 - x^3} \right) \\ & = \left[ (2 + x^2)^3 (1 - x^3)^4 \right] \left( {6x \over 2 + x^2} - {12x^2 \over 1 - x^3} \right) \\ & = 6x(2 + x^2)^2 (1 - x^3)^4 - 12x^2 (2 + x^2)^3 (1 - x^3)^3 \\ & = 6x(2 + x^2)^2 (1 - x^3)^3 [ (1 - x^3) - 2x (2 + x^2) ] \\ & = 6x(2 + x^2)^2 (1 - x^3)^3 (1 - x^3 - 4x - 2x^3) \\ & = 6x(2 + x^2)^2 (1 - x^3)^3 (1 - 4x - 3x^3) \end{align}$$

Note: I don't think this will be tested in O levels (more for A levels)

(i)

$$\begin{align} y & = { \left(x + {1 \over x}\right)^5 (5 + x) e^x \over 10^x \ln x } \\ \ln y & = \ln \left[ \left(x + {1 \over x}\right)^5 (5 + x) e^x \over 10^x \ln x \right] \\ \ln y & = \ln \left[ \left(x + {1 \over x}\right)^5 (5 + x) e^x \right] - \ln (10^x \ln x ) \\ \ln y & = \ln \left(x + {1 \over x}\right)^5 + \ln (5 + x) + \ln e^x - \ln 10^x - \ln (\ln x) \\ \ln y & = 5 \ln (x + x^{-1}) + \ln (5 + x) + x - x \ln 10 - \ln (\ln x) \\ \\ {1 \over y} {dy \over dx} & = 5 \left(1 - x^{-2} \over x + x^{-1} \right) + {1 \over 5 + x} + 1 - \ln 10 - { {1 \over x} \over \ln x } \\ {1 \over y} {dy \over dx} & = 5 \left( 1 - {1 \over x^2} \over x + {1 \over x} \right) + {1 \over 5 + x} + 1 - \ln 10 - { 1 \over x \ln x} \\ {1 \over y} {dy \over dx} & = {5 \left(1 - {1 \over x^2} \right) \over x + {1 \over x} } + {1 \over 5 + x} + 1 - \ln 10 - {1 \over x \ln x} \\ \\ {dy \over dx} & = y\left[ {5 \left(1 - {1 \over x^2} \right) \over x + {1 \over x} } + {1 \over 5 + x} + 1 - \ln 10 - {1 \over x \ln x} \right] \\ \end{align}$$

(ii)

$$\begin{align} y & = { (\cos^5 x) (x^2 + 5) (\ln x) \over e^{x + 1} } \\ \ln y & = \ln \left[ (\cos^5 x) (x^2 + 5) (\ln x) \over e^{x + 1} \right] \\ \ln y & = \ln [(\cos^5 x)(x^2 + 5)(\ln x)] - \ln e^{x + 1} \\ \ln y & = \ln (\cos^5 x) + \ln (x^2 + 5) + \ln (\ln x) - (x + 1) \\ \ln y & = 5 \ln (\cos x) + \ln (x^2 + 5) + \ln (\ln x) - x - 1 \\ \\ {1 \over y} {dy \over dx} & = 5 \left(- \sin x \over \cos x\right) + {2x \over x^2 + 5} + { {1 \over x} \over \ln x } - 1 \\ {1 \over y} {dy \over dx} & = - 5 \tan x + {2x \over x^2 + 5} + {1 \over x \ln x} - 1 \\ {dy \over dx} & = y \left( - 5 \tan x + {2x \over x^2 + 5} + {1 \over x \ln x} - 1 \right) \end{align}$$

(i)

$$\begin{align} y & = ax - x \ln x \\ \\ \text{Let } & y = 0, \\ 0 & = ax - x \ln x \\ 0 & = x (a - \ln x) \end{align}$$ $$\begin{align} x & = 0 \text{ (Reject, since } x > 0) && \text{ or } & a - \ln x & = 0 \\ & &&& - \ln x & = -a \\ & &&& \ln x & = a \\ & &&& \log_e x & = a \\ & &&& x & = e^a \end{align}$$

(ii)

$$\begin{align} u & = x &&& v & = \ln x \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over x} \end{align}$$ $$\begin{align} {dy \over dx} & = a(1) - \left[ (x) \left(1 \over x\right) + (\ln x)(1) \right] \\ & = a - (1 + \ln x) \\ & = a - 1 - \ln x \\ \\ {d^2 y \over dx^2 } & = - {1 \over x} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = a - 1 - \ln x \\ \ln x & = a -1 \\ \log_e x & = a - 1 \\ x & = e^{a - 1} \\ \\ \text{Substitute } & x = e^{a - 1} \text{ into eqn of curve,} \\ y & = a(e^{a - 1}) - (e^{a - 1}) \ln e^{a - 1} \\ & = a e^{a - 1} - e^{a - 1} (a - 1) \ln e \\ & = a e^{a - 1} - e^{a - 1} (a - 1) (1) \\ & = a e^{a - 1} - a e^{a - 1} + e^{a - 1} \\ & = e^{a - 1} \\ \\ \text{Coordinates } & \text{of stationary point is } (e^{a - 1}, e^{a - 1}) \\ \\ \text{Substitute } & x = e^{a - 1} \text{ into } {d^2 y \over dx^2} ,\\ {d^2 y \over dx^2 } & = - {1 \over e^{a - 1}} \\ \\ \text{Since } a > 0 & \text{ and } e^{a - 1} > 0, \phantom{.} {d^2 y \over dx^2} < 0 \\ \\ \therefore (e^{a - 1}, e^{a - 1}) & \text{ is a maximum point} \end{align}$$

(i) The dotted line denotes the portion of y = x ln x that was reflected about the x-axis to obtain the graph of y = |x ln x|

(ii)

$$\begin{align} \text{Consider } y & = x \ln x \\ \\ u & = x \phantom{00000000} {du \over dx} = 1 \\ v & = \ln x \phantom{000000} {dv \over dx} = {1 \over x} \\ \\ {dy \over dx} & = x \left(1 \over x\right) + (\ln x)(1) \\ & = 1 + \ln x \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 1 + \ln x \\ - \ln x & = 1 \\ \ln x & = -1 \\ \log_e x & = -1 \\ x & = e^{-1} \\ \\ \text{Substitute } & \text{into } y = x \ln x, \\ y & = e^{-1} \ln e^{-1} \\ y & = e^{-1} (-1) \\ y & = -e^{-1} \\ \\ \text{Minimum } & \text{point on } y = x \ln x : \left({1 \over e}, -{1 \over e} \right) \\ \\ \\ \text{Maximum } & \text{point on } y = |x \ln x| : \left({1 \over e}, {1 \over e} \right) \end{align}$$

$$\begin{align} \text{Tom: } {d \over dx}(x^n) & = nx^{n - 1} \\ \\ n \text{ is an integer, } & \text{not a variable} \\ \\ \\ \text{Doris: } {d \over dx} (a^x) & = a^x \ln a \\ \\ a \text{ is an integer, } & \text{not a variable} \\ \\ \\ \text{Let } y & = x^x \\ \ln y & = \ln x^x \\ \ln y & = x \ln x \\ {1 \over y}{dy \over dx} & = x\left(1 \over x\right) + (\ln x)(1) \\ {1 \over y}{dy \over dx} & = 1 + \ln x \\ {dy \over dx} & = y(1 + \ln x) \\ {dy \over dx} & = x^x (1 + \ln x) \end{align}$$

$$\begin{align} \text{Curve 1: } y & = \ln x \\ \\ \\ \text{Curve 2: } y & = \ln 3x \\ y & = \ln 3 + \ln x \\ y & = \ln x + \ln 3 \\ y & \approx \ln x + 1.10 \\ \\ \\ \text{Curve 2 is a } & \text{translation of curve 1 by approx. 1.10 units} \end{align}$$

$$\begin{align} {d \over dx} [ \ln |ax|] & = {d \over dx} [ \ln (|a||x|)] \\ & = {d \over dx} [ \ln |a| + \ln |x| ] \\ & = 0 + {d \over dx} [\ln |x|] \end{align}$$