(a)

\begin{align} {d \over dx} (e^x + 1) & = (1)e^x \\ & = e^x \end{align}

(b)

\begin{align} {d \over dx} (4e^x - 6) & = 4(1)e^x \\ & = 4e^x \end{align}

(c)

\begin{align} u & = x &&& v & = e^x \\ {du \over dx} & = 1 &&& {dv \over dx} & = (1)e^x \\ & &&& & = e^x \end{align} \begin{align} {d \over dx} (xe^x) & = (x)(e^x) + (e^x)(1) \phantom{000000} [\text{Product rule}] \\ & = xe^x + e^x \\ & = e^x (x + 1) \end{align}

(d)

\begin{align} u & = e^x &&& v & = x \\ {du \over dx} & = (1)e^x &&& {dv \over dx} & = 1 \\ & = e^x \end{align} \begin{align} {d \over dx} \left({e^x \over x}\right) & = {(x)(e^x) - (e^x)(1) \over x^2} \phantom{000000} [\text{Quotient rule}] \\ & = {xe^x - e^x \over x^2} \\ & = {e^x (x - 1) \over x^2} \end{align}

(a)

\begin{align} {d \over dx} (e^{-2x}) & = (-2)e^{-2x} \\ & = - 2e^{-2x} \end{align}

(b)

\begin{align} {d \over dx} (5e^{3 - 5x}) & = 5(-5)e^{3 - 5x} \\ & = -25 e^{3 - 5x} \end{align}

(c)

\begin{align} {d \over dx} (e^{3x} + 7) & = (3)e^{3x} \\ & = 3e^{3x} \end{align}

(d)

\begin{align} {d \over dx} (e^x + e^{-x}) & = (1)e^{x} + (-1)e^{-x} \\ & = e^x - e^{-x} \end{align}

(a)

\begin{align} y & = e^{3x} . e^{x - 1} \\ & = e^{3x + x - 1} \\ & = e^{4x - 1} \\ \\ {dy \over dx} & = (4)e^{4x - 1} \\ & = 4e^{4x - 1} \end{align}

(b)

\begin{align} y & = {e^{2x + 3} \over e^x} \\ & = e^{2x + 3 - x} \\ & = e^{x + 3} \\ \\ {dy \over dx} & = (1)e^{x + 3} \\ & = e^{x + 3} \end{align}

(c)

\begin{align} y & = (e^{1 - x})^2 \\ & = e^{2(1 - x)} \\ & = e^{2 - 2x} \\ \\ {dy \over dx} & = (-2)e^{2 - 2x} \\ & = - 2 e^{2 - 2x} \end{align}

(d)

\begin{align} y & = \sqrt{e^{3x}} \\ & = (e^{3x})^{1 \over 2} \\ & = e^{{3 \over 2}x} \\ \\ {dy \over dx} & = {3 \over 2} e^{{3 \over 2}x} \end{align}

Use the chain rule formula ${dy \over dx} = {dy \over du} \times {du \over dt}$ for the entire question

(a)

\begin{align} y & = (1 - e^x)^2 \\ \\ \text{Let } & u = 1 - e^x, \\ y & = u^2 \end{align} \begin{align} {du \over dx} & = {d \over dx} (1 - e^x) &&& {dy \over du} & = {d \over du} (u^2) \\ & = 0 - (1)e^x &&& & = 2u \\ & = -e^x \end{align} \begin{align} {dy \over dx} & = {dy \over du} \times {du \over dx} \\ & = 2u \times (-e^x) \\ & = 2(u)(-e^x) \\ & = 2(1 - e^x)(-e^x) \\ & = -2e^x(1 - e^x) \end{align}

(b)

\begin{align} y & = (x + e^x)^5 \\ \\ \text{Let } & u = x + e^x, \\ y & = u^5 \end{align} \begin{align} {du \over dx} & = {d \over dx} (x + e^x) &&& {dy \over du} & = {d \over du} (u^5) \\ & = 1 + (1)e^x &&& & = 5u^4 \\ & = 1 + e^x \end{align} \begin{align} {dy \over dx} & = {dy \over du} \times {du \over dx} \\ & = 5u^4 \times (1 + e^x) \\ & = 5(u)^4(1 + e^x) \\ & = 5(x + e^x)^4(1 + e^x) \end{align}

(c)

\begin{align} y & = \sqrt{1 + e^x} \\ \\ \text{Let } & u = 1 + e^x, \\ y & = \sqrt{u} \\ & = u^{1 \over 2} \end{align} \begin{align} {du \over dx} & = {d \over dx} (1 + e^x) &&& {dy \over du} & = {d \over du} (u^{1 \over 2}) \\ & = 0 + (1)e^x &&& & = {1 \over 2}u^{-{1 \over 2}} \\ & = e^x &&& & = {1 \over 2}\left({1 \over \sqrt{u}}\right) \\ & &&& & = {1 \over 2\sqrt{u}} \end{align} \begin{align} {dy \over dx} & = {dy \over du} \times {du \over dx} \\ & = {1 \over 2\sqrt{u}} \times (e^x) \\ & = {e^x \over 2\sqrt{u}} \\ & = {e^x \over 2\sqrt{1 + e^x}} \end{align}

(d)

\begin{align} y & = \sqrt{x + e^x} \\ \\ \text{Let } & u = x + e^x, \\ y & = \sqrt{u} \\ & = u^{1 \over 2} \end{align} \begin{align} {du \over dx} & = {d \over dx} (x + e^x) &&& {dy \over du} & = {d \over du} (u^{1 \over 2}) \\ & = 1 + (1)e^x &&& & = {1 \over 2}u^{-{1 \over 2}} \\ & = 1 + e^x &&& & = {1 \over 2}\left({1 \over \sqrt{u}}\right) \\ & &&& & = {1 \over 2\sqrt{u}} \end{align} \begin{align} {dy \over dx} & = {dy \over du} \times {du \over dx} \\ & = {1 \over 2\sqrt{u}} \times (1 + e^x) \\ & = {1 + e^x \over 2\sqrt{u}} \\ & = {1 + e^x \over 2\sqrt{x + e^x}} \end{align}

(a)

\begin{align} u & = x + 1 &&& v & = e^{2x} \\ {du \over dx} & = 1 &&& {dv \over dx} & = (2)e^{2x} \\ & &&& & = 2e^{2x} \end{align} \begin{align} {dy \over dx} & = (x + 1)(2e^{2x}) + (e^{2x})(1) \\ & = 2(x + 1)e^{2x} + e^{2x} \\ & = e^{2x} [2(x + 1) + 1] \\ & = e^{2x} (2x + 2 + 1) \\ & = e^{2x} (2x + 3) \end{align}

(b)

\begin{align} u & = e^{-2x} &&& v & = \sin x \\ {du \over dx} & = (-2)e^{-2x} &&& {dv \over dx} & = \cos x \\ & = -2 e^{-2x} \end{align} \begin{align} {dy \over dx} & = (e^{-2x})(\cos x) + (\sin x)(-2 e^{-2x}) \\ & = e^{-2x} \cos x - 2 e^{-2x} \sin x \\ & = e^{-2x} (\cos x - 2 \sin x) \end{align}

(c)

\begin{align} u & = x &&& v & = e^{x^3 + 2x} \\ {du \over dx} & = 1 &&& {dv \over dx} & = (3x^2 + 2)e^{x^3 + 2x} \end{align} \begin{align} {dy \over dx} & = (x)[(3x^2 + 2)e^{x^3 + 2x}] + (e^{x^3 + 2x})(1) \\ & = x(3x^2 + 2)e^{x^3 + 2x} + e^{x^3 + 2x} \\ & = e^{x^3 + 2x} [x(3x^2 + 2) + 1] \\ & = e^{x^3 + 2x} (3x^3 + 2x + 1) \end{align}

(d)

\begin{align} u & = e^{1 - x} &&& v & = \cos 2x \\ {du \over dx} & = (-1)e^{1 - x} &&& {dv \over dx} & = (2)(-\sin 2x) \\ & = -e^{1 - x} &&& & = -2 \sin 2x \end{align} \begin{align} {dy \over dx} & = (e^{1 - x})(-2 \sin 2x) + (\cos 2x)(-e^{1 - x}) \\ & = - 2e^{1 - x} \sin 2x - e^{1 - x} \cos 2x \\ & = - e^{1 - x} (2 \sin 2x + \cos 2x) \end{align}

(a)

\begin{align} u & = x &&& v & = 1 + e^{-x} \\ {du \over dx} & = 1 &&& {dv \over dx} & = (-1)e^{-x} \\ & &&& & = -e^{-x} \end{align} \begin{align} {dy \over dx} & = {(1 + e^{-x})(1) - (x)(-e^{-x}) \over (1 + e^{-x})^2} \\ & = {1 + e^{-x} + x e^{-x} \over (1 + e^{-x})^2 } \end{align}

(b)

\begin{align} u & = e^x &&& v & = 1 + e^{2x} \\ {du \over dx} & = (1)e^x &&& {dv \over dx} & = (2)e^{2x} \\ & = e^x &&& & = 2e^{2x} \end{align} \begin{align} {dy \over dx} & = { (1 + e^{2x})(e^x) - (e^x)(2e^{2x}) \over (1 + e^{2x})^2} \\ & = {e^x (1 + e^{2x}) - 2 e^x (e^{2x}) \over (1 + e^{2x})^2} \\ & = {e^x [(1 + e^{2x}) - 2e^{2x}] \over (1 + e^{2x})^2} \\ & = {e^x (1 - e^{2x}) \over (1 + e^{2x})^2} \end{align}

(c)

\begin{align} u & = e^{3x} &&& v & = \cos x \\ {du \over dx} & = (3)e^{3x} &&& {dv \over dx} & = -\sin x \\ & = 3e^{3x} \end{align} \begin{align} {dy \over dx} & = {(\cos x)(3e^{3x}) - (e^{3x})(-\sin x) \over (\cos x)^2} \\ & = { 3e^{3x} \cos x + e^{3x} \sin x \over \cos^2 x} \\ & = { e^{3x} (3 \cos x + \sin x) \over \cos^2 x} \end{align}

(d)

\begin{align} u & = x + e^{2x} &&& v & = e^x \\ {du \over dx} & = 1 + (2)e^{2x} &&& {dv \over dx} & = (1)e^x \\ & = 1 + 2e^{2x} &&& & = e^x \end{align} \begin{align} {dy \over dx} & = {(e^x)(1 + 2e^{2x}) - (x + e^{2x})(e^x) \over (e^x)^2} \\ & = {e^x [ (1 + 2e^{2x}) - (x + e^{2x}) ] \over (e^x)^2} \\ & = {e^x (1 + 2e^{2x} - x - e^{2x} ) \over (e^x)^2 } \\ & = {e^x (1 + e^{2x} - x) \over (e^x)^2 } \\ & = {1 + e^{2x} - x \over e^x} \end{align}

\begin{align} y & = 2e^x + 1 \\ \\ {dy \over dx} & = 2(1)e^x \\ & = 2e^x \\ \\ {d^2 y \over dx^2} & = 2(1)e^x \\ & = 2e^x \\ \\ \text{Substitute } & x = 1 \text{ into } {dy \over dx}, \\ {dy \over dx} & = 2e^{(1)} \\ & = 2e \\ \\ \text{Substitute } & x = 1 \text{ into } {d^2 y \over dx^2}, \\ {d^2y \over dx^2} & = 2e^{(1)} \\ & = 2e \end{align}

(i)

\begin{align} y & = 3e^{2x} \\ \\ \text{When } & y = 3, \\ 3 & = 3e^{2x} \\ 1 & = e^{2x} \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln 1 & = \ln e^{2x} \\ 0 & = \ln e^{2x} \\ 0 & = 2x \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ 0 & = 2x (1) \\ 0 & = 2x \\ 0 & = x \end{align}

(ii)

\begin{align} y & = 3e^{2x} \\ \\ {dy \over dx} & = 3(2)e^{2x} \\ & = 6e^{2x} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = 6e^{2(1)} \\ & = 6e^2 \end{align}

(iii)

\begin{align} {dy \over dx} & = 6e^{2x} \\ \\ {d^2 y \over dx^2} & = 6(2)e^{2x} \\ & = 12e^{2x} \\ \\ \text{Let } & {dy \over dx} = 3, \\ 3 & = 6e^{2x} \\ {3 \over 6} & = e^{2x} \\ 0.5 & = e^{2x} \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln 0.5 & = \ln e^{2x} \\ \ln 0.5 & = 2x \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln 0.5 & = 2x (1) \\ \ln 0.5 & = 2x \\ {\ln 0.5 \over 2} & = x \\ \\ \text{Substitute } & x = {\ln 0.5 \over 2} \text{ into } {d^2 y \over dx^2}, \\ {d^2y \over dx^2} & = 12e^{2({\ln 0.5 \over 2})} \\ & = 12e^{\ln 0.5} \\ & = 12 (e^{\ln 0.5}) \\ & = 12(0.5) \\ & = 6 \end{align}

(i)

\begin{align} y & = {e^{-2x} \over 6} \\ & = {1 \over 6}e^{-2x} \\ \\ {dy \over dx} & = {1 \over 6}(-2)e^{-2x} \\ & = -{1 \over 3}e^{-2x} \end{align}

(ii)

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

Note: E, C and R are constants while Q and t are variables

\begin{align} Q & = -EC e^{-{t \over RC}} \\ \\ {dQ \over dt} & = -EC \left(- {1 \over RC} \right) e^{-{t \over RC}} \\ & = {EC \over RC} \left( e^{-{t \over RC}} \right) \\ & = {E \over R}e^{-{t \over RC}} \\ \\ \text{When } & t = 3, \\ {dQ \over dt} & = {E \over R}e^{-{3 \over RC}} \end{align}

(a)

\begin{align} {d \over dx} (e^{2x - 1} + x)^3 & = 3(e^{2x - 1} + x)^2 . [(2)e^{2x - 1} + 1] \phantom{000000} [\text{Chain rule}] \\ & = 3(e^{2x - 1} + x)^2 (2e^{2x - 1} + 1) \end{align}

(b)

\begin{align} u & = 2 + e^{1 - x} &&& v & = \tan 2x \\ {du \over dx} & = (-1)e^{1 - x} &&& {dv \over dx} & = (2)\sec^2 2x \\ & = - e^{1 - x} &&& & = 2 \sec^2 2x \end{align} \begin{align} {d \over dx} \left( 2 + e^{1 - x} \over \tan 2x \right) & = {(\tan 2x)(-e^{1 - x}) - (2 + e^{1 - x})(2 \sec^2 2x) \over (\tan 2x)^2} \phantom{000000} [\text{Quotient rule}] \\ & = { - e^{1 - x} \tan 2x - 2(2 + e^{1 - x}) \sec^2 2x \over \tan^2 2x } \end{align}

(c)

\begin{align} u & = x &&& v & = \cos x \\ {du \over dx} & = 1 &&& {dv \over dx} & = - \sin x \end{align} \begin{align} {d \over dx} (x \cos x) & = (x)(- \sin x) + (\cos x)(1) \phantom{000000} [\text{Product rule}] \\ & = -x \sin x + \cos x \end{align}
\begin{align} u & = x \cos x &&& v & = e^{1 - x^2} \\ {du \over dx} & = -x \sin x + \cos x &&& {dv \over dx} & = (-2x)e^{1 - x^2} \\ & &&& & = - 2x e^{1 - x^2} \end{align} \begin{align} {d \over dx} \left(x \cos x \over e^{1 - x^2} \right) & = {e^{1 - x^2} (-x \sin x + \cos x) - (x \cos x)(-2x e^{1 - x^2}) \over (e^{1 - x^2})^2 } \phantom{00000} [\text{Quotient rule}] \\ & = {e^{1 - x^2} (-x \sin x + \cos x) + 2x^2 e^{1 - x^2} \cos x \over (e^{1 - x^2})^2 } \\ & = {e^{1 - x^2} [ (-x \sin x + \cos x) + 2x^2 \cos x] \over (e^{1 - x^2})^2 } \\ & = { - x \sin x + \cos x + 2x^2 \cos x \over e^{1 -x^2} } \\ & = { - x \sin x + \cos x(1 + 2x^2) \over e^{1 - x^2} } \end{align}

(i)

\begin{align} y & = k \left( e^{x \over 2k} + e^{-{x \over 2k}} \right) \\ \\ \text{When } & k = {1 \over 2}, \\ y & = \left(1 \over 2\right) \left[ e^{x \over 2({1 \over 2})} + e^{-{x \over 2({1 \over 2})}} \right] \\ & = {1 \over 2} \left( e^{x \over 1} + e^{-x \over 1} \right) \\ & = {1 \over 2} \left( e^x + e^{-x} \right) \\ & = { e^x + e^{-x} \over 2} \end{align}

(ii)

\begin{align} y & = {1 \over 2} \left(e^x + e^{-x} \right) \\ & = {1 \over 2}e^x + {1 \over 2}e^{-x} \\ \\ {dy \over dx} & = {1 \over 2}(1)e^x + {1 \over 2}(-1)e^{-x} \\ & = {1 \over 2}e^x - {1 \over 2}e^{-x} \\ \\ {d^2 y \over dx^2} & = {1 \over 2}(1)e^x - {1 \over 2}(-1)e^{-x} \\ & = {1 \over 2}e^x + {1 \over 2}e^{-x} \end{align}

(i) Recall that the tangent & normal to the curve at the same point are perpendicular, i.e. $m_1 \times m_2 = -1$

\begin{align} y & = e^{2x - 1} \\ \\ \text{When } & x = 2, \\ y & = e^{2(2) - 1} \\ & = e^3 \\ \\ \therefore & \phantom{.} A(2, e^3) \\ \\ y & = e^{2x - 1} \\ \\ {dy \over dx} & = (2)e^{2x - 1} \\ & = 2e^{2x - 1} \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = 2e^{2(2) - 1} \\ & = 2e^3 \\ \\ y & = mx + c \\ y & = 2e^3 x + c \\ \\ \text{Using } & A(2, e^3), \\ e^3 & = 2e^3 (2) + c \\ e^3 & = 4e^3 + c \\ - 3 e^3 & = c \\ \\ \text{Eqn of tangent: } & y = 2e^3 x - 3e^3 \\ y & = e^3 (2x - 3) \\ \\ \\ \text{Gradient of normal} & = {-1 \over 2e^3} \\ & = -{1 \over 2e^3} \\ \\ y & = mx + c \\ y & = -{1 \over 2e^3} x + c \\ \\ \text{Using } & A(2, e^3), \\ e^3 & = -{1 \over 2e^3} (2) + c \\ e^3 & = - {2 \over 2e^3} + c \\ e^3 & = -{1 \over e^3} + c \\ e^3 + {1 \over e^3} & = c \\ \\ \text{Eqn of normal: } & y = -{1 \over 2e^3} x + e^3 + {1 \over e^3} \end{align}

(ii) Since the points B and C lie on the x-axis, the height of the triangle is the vertical distance between A and the x-axis.

\begin{align} \text{Eqn of tangent: } & y = 2e^3 x - 3e^3 \\ \\ \text{Let } & y = 0, \\ 0 & = 2e^3 x - 3e^3 \\ 3e^3 & = 2e^3 x \\ {3e^3 \over 2e^3} & = x \\ {3 \over 2} & = x \\ \\ \therefore & \phantom{.} B \left({3 \over 2}, 0\right) \\ \\ \\ \text{Eqn of normal: } & y = -{1 \over 2e^3} x + e^3 + {1 \over e^3} \\ \\ \text{Let } & y = 0, \\ 0 & = -{1 \over 2e^3} x + e^3 + {1 \over e^3} \\ {1 \over 2e^3} x & = e^3 + {1 \over e^3} \\ x & = 2e^3 \left(e^3 + {1 \over e^3}\right) \\ & = 2e^6 + {2e^3 \over e^3} \\ & = 2e^6 + 2 \\ \\ \therefore & \phantom{.} C(2e^6 + 2, 0) \\ \\ BC & = 2e^6 + 2 - {3 \over 2} \\ & = 2e^6 + {1 \over 2} \\ & = {4e^6 \over 2} + {1 \over 2} \\ & = {4e^6 + 1 \over 2} \\ \\ \text{Area of triangle } ABC & = {1 \over 2} \times BC \times \text{Height} \\ & = {1 \over 2} \times {4e^6 + 1 \over 2} \times e^3 \\ & = {e^3 (4e^6 + 1) \over 4} \end{align}

(i)

\begin{align} \text{Curve 1: } & y = e^{2x - 3} \\ \\ \text{When } & x = 2, \\ y & = e^{2(2) - 3} \\ & = e \\ \\ \therefore P & \phantom{.} (2, e) \\ \\ \text{Curve 2: } & y = e^{a - x} \\ \\ \text{Using } & P(2, e), \\ e & = e^{a - 2} \\ e^1 & = e^{a - 2} \\ \\ \therefore 1 & = a - 2 \\ 3 & = a \end{align}

(ii)

\begin{align} \text{Curve 1: } & y = e^{2x - 3} \\ \\ {dy \over dx} & = (2)e^{2x - 3} \\ & = 2e^{2x - 3} \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = 2e^{2(2) - 3} \\ & = 2e \\ \\ \\ \text{Curve 2: } & y = e^{3 - x} \\ \\ {dy \over dx} & = (-1)e^{3 - x} \\ & = - e^{3 - x} \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = - e^{3 - 2} \\ & = -e \end{align}

(i)

\begin{align} u & = 5x &&& v & = e^{-x} \\ {du \over dx} & = 5 &&& {dv \over dx} & = (-1)e^{-x} \\ & &&& & = -e^{-x} \end{align} \begin{align} {dy \over dx} & = (5x)(-e^{-x}) + (e^{-x})(5) \phantom{000000} [\text{Product rule}] \\ & = - 5x e^{-x} + 5 e^{-x} \\ & = 5e^{-x} (- x + 1) \end{align}
\begin{align} u & = 5e^{-x} &&& v & = - x+ 1 \\ {du \over dx} & = 5(-1)e^{-x} &&& {dv \over dx} & = -1 \\ & = - 5 e^{-x} \end{align} \begin{align} {d^2 y \over dx^2} & = (5e^{-x})(-1) + (-x + 1)(-5 e^{-x}) \\ & = - 5e^{-x} - 5e^{-x} (-x + 1) \\ & = 5e^{-x} [ - 1 - (-x + 1)] \\ & = 5e^{-x} ( - 1 + x - 1 ) \\ & = 5e^{-x} (x - 2) \end{align}

(ii)

\begin{align} {dy \over dx} & = 5e^{-x} (- x + 1) \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 5e^{-x} (- x + 1) \end{align} \begin{align} 5e^{-x} & = 0 && \text{ or } & -x + 1 & = 0 \\ e^{-x} & = 0 \text{ (Reject, since } e^{-x} > 0) &&& 1 & = x \end{align}
$$ \therefore x = 1 $$

(iii)

\begin{align} {d^2 y \over dx^2} & = 5e^{-x} (x - 2) \\ \\ \text{When } & x = 1, \\ {d^2 y \over dx^2} & = 5e^{-1} (1 - 2)\\ & = 5\left(1 \over e\right) (-1) \\ & = -{5 \over e} \end{align}

(i)

\begin{align} y & = e^x - 2x - 1 \\ \\ {dy \over dx} & = (1)e^x - 2(1) \\ & = e^x - 2 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = e^x - 2 \\ 2 & = e^x \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln 2 & = \ln e^x \\ \ln 2 & = x \ln e \phantom{00000000} [\text{Power law (logarithms)}] \\ \ln 2 & = x (1) \\ \ln 2 & = x \end{align}

(ii)

\begin{align} y & = e^x - 2x - 1 \\ \\ \text{When } & x = \ln 2, \\ y & = e^{(\ln 2)} - 2(\ln 2) - 1 \\ & = (2) - 2\ln 2 - 1 \\ & = 1 - 2\ln 2 \\ & = -0.38629 \\ & \approx -0.386 \end{align}

(iii) $y$ is increasing for the region to the right of the minimum point, $(\ln 2, 1 - 2\ln 2)$

$$ x > \ln 2 $$

(i)

\begin{align} y & = e^{-x^2 + 2x} \\ \\ {dy \over dx} & = (-2x + 2) e^{-x^2 + 2x} \\ & = -2(x - 1) e^{-x^2 + 2x} \end{align}

(ii)

\begin{align} {dy \over dx} & = -2(x - 1) e^{-x^2 + 2x} \\ \\ \text{Let } & {dy \over dx} = 0 ,\\ 0 & = -2(x - 1) e^{-x^2 + 2x} \end{align} \begin{align} x - 1 & = 0 && \text{ or } & e^{-x^2 + 2x} & = 0 \text{ (Reject, since } e^{-x^2 + 2x} > 0 ) \\ x & = 1 \\ \\ \text{Substitute } & \text{into eqn of curve,} \\ y & = e^{-(1)^2 + 2(1)} \\ & = e \phantom{00} \text{ (Shown)} \end{align}

(iii)

(i)

\begin{align} P & = 650 \phantom{.} 000 e^{mt} \\ \\ \text{When } & t = 0, \\ P & = 650 \phantom{.} 000e^{m(0)} \\ & = 650 \phantom{.} 000(1) \\ & = 650 \phantom{.} 000 \end{align}

(ii)

\begin{align} P & = 650 \phantom{.} 000 e^{mt} \\ \\ {dP \over dt} & = 650 \phantom{.} 000 (m) e^{mt} \\ & = (650 \phantom{.} 000 e^{mt})(m) \\ & = (P)(m) \\ & = Pm \\ \\ \text{Since } m & \text{ is a constant,} \\ {dP \over dt} & \varpropto P \end{align}

(iii)

\begin{align} P & = 650 \phantom{.} 000 e^{mt} \\ \\ \text{When } t = 25 & \text{ and } P = 995 \phantom{.} 000, \\ 995 \phantom{.} 000 & = 650 \phantom{.} 000e^{m(25)} \\ 995 \phantom{.} 000 & = 650 \phantom{.} 000e^{25m} \\ {995 \phantom{.} 000 \over 650 \phantom{.} 000} & = e^{25m} \\ {199 \over 130} & = e^{25m} \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln{199 \over 130} & = \ln e^{25m} \\ \ln{199 \over 130} & = 25m (\ln e) \phantom{00000000000} [\text{Power law (logarithms)}] \\ \ln{199 \over 130} & = 25m (1) \\ \ln {199 \over 130} & = 25m \\ \\ 25m & = \ln {199 \over 130} \\ 25m & = \ln 199 - \ln 130 \phantom{000000} [\text{Quotient law (logarithms)}] \\ m & = {\ln 199 - \ln 130 \over 25} \end{align}

(iv)

\begin{align} P & = 650 \phantom{.} 000 e^{mt} \\ \\ \text{When } & t = 30, \\ P & = 650 \phantom{.} 000e^{{\ln 199 - \ln 130 \over 25}(30)} \\ & = 1 \phantom{.} 083 \phantom{.} 440 \\ & \approx 1 \phantom{.} 083 \phantom{.} 000 \end{align}

(v) From 2006 to 2014, 8 years has passed.

\begin{align} P & = 650 \phantom{.} 000 e^{mt} \\ \\ \text{When } & t = 8, \\ P & = 650000e^{{\ln 199 - \ln 130 \over 25}(8)} \\ & = 744 \phantom{.} 876 \\ \\ \text{When } & t = 9, \\ P & = 650000e^{{\ln 199 - \ln 130 \over 25}(9)} \\ & = 757 \phantom{.} 671 \\ \\ \text{Increase in population } & = 757 \phantom{.} 671 - 744 \phantom{.} 876 \\ & = 12 \phantom{.} 795 \\ & \approx 13 \phantom{.} 000 \end{align}

(vi)

\begin{align} {dP \over dt} & > 18000 \\ (650000e^{mt})(m) & > 18000 \\ 650000e^{mt} & > {18000 \over m} \\ e^{mt} & > {18000 \over 650000m} \\ e^{mt} & > {9 \over 325m} \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln e^{mt} & > \ln {9 \over 325m} \\ [\text{Power law (logarithms)}] \phantom{000000} mt (\ln e) & > \ln {9 \over 325m} \\ mt (1) & > \ln {9 \over 325m} \\ mt & > \ln {9 \over 325m} \\ t & > {\ln {9 \over 325m} \over m} \\ \\ \text{When } & m = {\ln 199 - \ln 130 \over 25}, \\ t & > 28.544 \\ \\ \therefore t & = 29 \end{align}

(i)

\begin{align} m & = ae^{-kt} \\ \\ \text{When } t = 0 & \text{ and } m = 100, \\ 100 & = ae^{-k(0)} \\ 100 & = a(e^0) \\ 100 & = a(1) \\ 100 & = a \end{align}

(ii)

\begin{align} m & = ae^{-kt} \\ & = 100e^{-kt} \\ \\ \text{When } t = 40 & \text{ and } m = 90, \\ 90 & = 100e^{-k(40)} \\ 90 & = 100e^{-40k} \\ {90 \over 100} & = e^{-40k} \\ {9 \over 10} & = e^{-40k} \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln {9 \over 10} & = \ln e^{-40k} \\ \ln {9 \over 10} & = (-40k)(\ln e) \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {9 \over 10} & = (-40k)(1) \\ \ln {9 \over 10} & = -40k \\ \\ k & = -{\ln {9 \over 10} \over 40} \\ & = -{1 \over 40} \ln {9 \over 10} \end{align}

(iii)

\begin{align} \text{Half of original mass} & = {1 \over 2} \times 100 \\ & = 50 \text{ g} \\ \\ m & = 100e^{-kt} \\ \\ 50 & = 100e^{-kt} \\ {50 \over 100} & = e^{-kt} \\ 0.5 & = e^{-kt} \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln 0.5 & = \ln e^{-kt} \\ \ln 0.5 & = (-kt)(\ln e) \phantom{000000} [\text{Power law (logarithms)}] \\ \ln 0.5 & = (-kt)(1) \\ \ln 0.5 & = -kt \\ \\ t & = -{\ln 0.5 \over k} \\ & = -{\ln 0.5 \over -{1 \over 40} \ln {9 \over 10}} \\ & = 263.15 \\ & \approx 263 \text{ h} \end{align}

(iv)

\begin{align} m & = 100e^{-kt} \\ \\ {dm \over dt} & = 100(-k)e^{-kt} \\ & = -100k e^{-kt} \\ \\ \text{When } & k = -{1 \over 40}\ln {9 \over 10}, \\ {dm \over dt} & = -100\left( -{1 \over 40} \ln {9 \over 10} \right) e^{-(-{1 \over 40} \ln {9 \over 10})t} \\ & = \left({5 \over 2} \ln {9 \over 10}\right)e^{({1 \over 40} \ln {9 \over 10})t} \end{align}

(i)

\begin{align} n & = 20e^{0.1t} \\ \\ \text{When } & t = 5, \\ n & = 20e^{0.1(5)} \\ & = 32.974 \\ & \approx 32 \end{align}

(ii)

\begin{align} n & > 200 \\ 20e^{0.1t} & > 200 \\ e^{0.1t} & > 10 \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln e^{0.1t} & > \ln 10 \\ [\text{Power law (logarithms)}] \phantom{000000} (0.1t)(\ln e) & > \ln 10 \\ (0.1t)(1) & > \ln 10 \\ 0.1t & > \ln 10 \\ t & > 10\ln 10 \\ t & > 23.025 \\ \\ \therefore \text{Least number of weeks} & = 24 \end{align}

(iii)

\begin{align} n & = 20e^{0.1t} \\ \\ {dn \over dt} & = 20(0.1)e^{0.1t} \\ & = 2e^{0.1t} \\ \\ \text{When } & t = 15, \\ {dn \over dt} & = 2e^{0.1(15)} \\ & = 8.9633 \\ & \approx 9 \text{ snails/week} \end{align}

(iv) Since $t > 0$ and $e^{0.1t}$ increases as $t$ increases, as $t$ increases, $n$ increases at an increasing rate

\begin{align} n & = 20e^{0.1t} \\ \\ \text{When } & t = 0, \\ n & = 20e^{0.1(0)} \\ & = 20(e^0) \\ & = 20(1) \\ & = 20 \\ \\ \text{Vertical-} & \text{intercept is } (0, 20) \end{align}

\begin{align} f(x) & = a(1 - e^{-bx}) \\ & = a - ae^{-bx} \\ \\ f'(x) & = - a(-b)e^{-bx} \\ & = ab e^{-bx} \\ \\ \text{For all real values} & \text{ of } x, \phantom{.} e^{-bx} > 0 \\ \\ \text{Since } a > 0 \text{ and } & b > 0, \phantom{.} f'(x) > 0 \\ \\ \therefore f(x) \text{ is increasing} & \text{ for all real values of } x \end{align}

\begin{align} y & = a e^{2x} + b e^{4x} \\ \\ {dy \over dx} & = a(2)e^{2x} + b (4) e^{4x} \\ & = 2a e^{2x} + 4b e^{4x} \\ \\ {d^2 y \over dx^2} & = 2 a (2) e^{2x} + 4b (4) e^{4x} \\ & = 4a e^{2x} + 16b e^{4x} \\ \\ \text{To show } & {d^2 y \over dx^2} = 6 {dy \over dx} - 8 y, \\ \text{R.H.S} & = 6 {dy \over dx} - 8 y \\ & = 6(2a e^{2x} + 4b e^{4x}) - 8(a e^{2x} + b e^{4x}) \\ & = 12a e^{2x} + 24b e^{4x} - 8ae^{2x} - 8b e^{4x} \\ & = 4a e^{2x} + 16be^{4x} \\ & = {d^2 y \over dx^2} \\ & = \text{L.H.S} \end{align}

\begin{align} \text{Curve 1: } & y = (2 + 3x)e^{-3x} \end{align} \begin{align} u & = 2 + 3x &&& v & = e^{-x} \\ {du \over dx} & = 3 &&& {dv \over dx} & = (-1)e^{-x} \\ & &&& & = -e^{-x} \end{align} \begin{align} {dy \over dx} & = (2 + 3x)(-e^{-x}) + (e^{-x})(3) \phantom{000000} [\text{Product rule}] \\ & = - e^{-x} (2 + 3x) + 3e^{-x} \\ & = - e^{-x} [ (2 + 3x) - 3 ] \\ & = - e^{-x} (2 + 3x - 3) \\ & = - e^{-x} (3x - 1) \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = -e^{-x} (3x - 1) \end{align} \begin{align} e^{-x} & = 0 \text{ (Reject, since } e^{-x} > 0) && \text{ or } & 3x - 1 & = 0 \\ & &&& 3x & = 1 \\ & &&& x & = {1 \over 3} \\ \\ & &&& \text{Substitute } & \text{into eqn of curve,} \\ & &&& y & = \left[ 2 + 3\left(1 \over 3\right) \right] e^{-{1 \over 3}} \\ & &&& & = (3)e^{-{1 \over 3}} \\ & &&& & = 3e^{-{1 \over 3}} \\ \\ & &&& \therefore & \phantom{.} \left({1 \over 3}, 3e^{-{1 \over 3}} \right) \end{align}
\begin{align} u & = -e^{-x} &&& v & = 3x - 1 \\ {du \over dx} & = - (-1)e^{-x} &&& {dv \over dx} & = 3 \\ & = e^{-x} \end{align} \begin{align} {d^2 y \over dx^2} & = (-e^{-x})(3) + (3x - 1)(e^{-x}) \\ & = - 3e^{-x} + e^{-x} (3x - 1) \\ & = e^{-x} [ - 3 + (3x - 1) ] \\ & = e^{-x} ( - 3 + 3x - 1 ) \\ & = e^{-x} (3x - 4) \\ \\ \text{When } & x = {1 \over 3}, \\ {d^2 y \over dx^2} & = e^{-{1 \over 3}} \left[ 3 \left(1 \over 3\right) - 4\right] \\ & = -2.1495 \\ \\ \therefore \left({1 \over 3}, 3e^{-{1 \over 3}} \right) & \phantom{.} \text{ is a max. point on Curve 1} \\ \\ \\ \text{Curve 2: } & y = 3e^{-x} \\ \\ \text{When } & x = {1 \over 3}, \\ y & = 3e^{-{1 \over 3}} \\ \\ \therefore \left({1 \over 3}, 3e^{-{1 \over 3}} \right) & \text{ lies on Curve 2} \end{align}

\begin{align} \text{Consider } y & = e^x - x \\ \\ {dy \over dx} & = e^x - 1 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = e^x - 1 \\ 1 & = e^x \\ e^0 & = e^x \\ 0 & = x \\ \\ \text{Substitute } & x = 0 \text{ into } y = e^x - x, \\ y & = e^0 - 0 \\ y & = 1 \\ \\ \\ {d^2 y \over dx^2} & = e^x \\ \\ \text{Substitute } & x = 0, \\ {d^2 y \over dx^2} & = e^0 \\ & = 1 > 0 \\ \\ \text{Minimum } & \text{point: } (0, 1) \\ \\ \\ \text{Since } (0, 1) \text{ is above } & x \text{-axis,} \text{ curve of } y = e^x - x \text{ does not meet } x \text{-axis} \\ \\ \\ \therefore e^x - x = 0 & \text{ has no solutions} \end{align}

\begin{align} y & = e^{kx} \\ \\ {dy \over dx} & = k e^{kx} \\ \\ {d^2 y \over dx^2} & = k (ke^{kx}) \\ & = k^2 e^{kx} \\ \\ {d^3 y \over dx^3} & = k^2 (k e^{kx}) \\ & = k^3 e^{kx} \\ & . \\ & . \\ & . \\ \\ {d^n y \over dx^n} & = k^n e^{kx} \end{align}

\begin{align} \text{Wrong technique: } & {d \over dx} (x^n) = n x^{n - 1} \\ \\ \text{Correct technique: } & {d \over dx} (e^x) = e^x \end{align}