By the product and chain rules, \[\frac{dy}{dx}=\frac32\sec^2\!\left(\frac x2\right)\cos2x-6\tan\!\left(\frac x2\right)\sin2x.\] At \(x=\pi/3\), \(\sec^2(\pi/6)=4/3\), \(\cos(2\pi/3)=-1/2\), \(\tan(\pi/6)=1/\sqrt3\) and \(\sin(2\pi/3)=\sqrt3/2\). Therefore \[\frac{dy}{dx}=\frac32\frac43\left(-\frac12\right)-6\frac1{\sqrt3}\frac{\sqrt3}{2}=-1-3=-4.\]

Differentiate implicitly with respect to \(x\), using the product and chain rules: \[2x\cos y-(x^2-3)\sin y\frac{dy}{dx}-12\cos^2y\sin y\frac{dy}{dx}=0.\]

At \(x=2,y=\pi/3\), \[2-\frac{\sqrt3}{2}\frac{dy}{dx}-\frac{3\sqrt3}{2}\frac{dy}{dx}=0.\] Thus \[2=2\sqrt3\frac{dy}{dx},\qquad\boxed{\frac{dy}{dx}=\frac1{\sqrt3}=\frac{\sqrt3}{3}}.\]

At \(P\), \[4(1+\ln x)=10+2\ln x\quad\Rightarrow\quad\ln x=3,\qquad P=(e^3,4).\]

By the quotient rule, \[\frac{dy}{dx}=\frac{(2/x)(1+\ln x)-(10+2\ln x)(1/x)}{(1+\ln x)^2}=-\frac8{x(1+\ln x)^2}.\] At \(P\), the gradient is \(-1/(2e^3)\). The tangent is \[y-4=-\frac1{2e^3}(x-e^3).\]

At \(Q\), \(y=0\), so \[-4=-\frac1{2e^3}(x_Q-e^3)\quad\Rightarrow\quad\boxed{x_Q=9e^3}.\]

(a) \[\frac{dx}{dt}=4e^{2t},\qquad\frac{dy}{dt}=\frac{e^t(9+e^{2t})-e^t(2e^{2t})}{(9+e^{2t})^2}=\frac{9e^t-e^{3t}}{(9+e^{2t})^2}.\] Therefore \[\boxed{\frac{dy}{dx}=\frac{9-e^{2t}}{4e^t(9+e^{2t})^2}}.\]

(b) The denominator is positive, so the stationary point occurs when \(e^{2t}=9\), giving \(e^t=3\). The derivative changes from positive to negative there, confirming a maximum. Substitution gives \[x=1+2(9)=19,\qquad y=\frac3{9+9}=\frac16,\qquad\boxed{M=(19,1/6)}.\]

(c) \(33=1+2e^{2t}\) gives \(e^{2t}=16\) and \(e^t=4\). Hence \[\frac{dy}{dx}=\frac{9-16}{4(4)(9+16)^2}=\boxed{-\frac7{10000}}=\boxed{-0.0007}.\]

At the \(y\)-axis, \(x=0\), so \(P=(0,\frac13)\). Apply the quotient rule:

\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2e^{2x}(e^x+2)-e^{2x}e^x}{(e^x+2)^2}=\frac{e^{3x}+4e^{2x}}{(e^x+2)^2}.\]

At \(x=0\), the tangent gradient is \(5/9\). The normal gradient is its negative reciprocal, \(-9/5\).

\[y-\frac13=-\frac95x\quad\Longrightarrow\quad\boxed{27x+15y-5=0}.\]

(a) Differentiate each parametric equation: \[\frac{\mathrm{d}x}{\mathrm{d}t}=6-2\cos2t,\qquad\frac{\mathrm{d}y}{\mathrm{d}t}=-6\sin2t+5\cos t.\]

At \(t=0\), these derivatives are 4 and 5 respectively. Hence \[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}=\boxed{\frac54}.\]

(b) Since \(\mathrm{d}x/\mathrm{d}t\ge4>0\), a stationary point of \(y\) as a function of \(x\) occurs when \(\mathrm{d}y/\mathrm{d}t=0\). Use \(\sin2t=2\sin t\cos t\):

\[-12\sin t\cos t+5\cos t=0\ \Longrightarrow\ \cos t(5-12\sin t)=0.\]

The interval excludes \(\pi/2\), so \(\cos t>0\) and \(\sin t=5/12\). Write \(s=\sin t\) and use \(\cos2t=1-2s^2\):

\[y=3-6s^2+5s=\frac{97}{24}-6\left(s-\frac5{12}\right)^2.\]

This is maximised at \(s=5/12\), which lies in the allowed range. Thus the exact \(y\)-coordinate of \(M\) is \(\boxed{97/24}\).

(a) \[\frac{dx}{d\theta}=\sec^2\theta,\qquad\frac{dy}{d\theta}=\cos\theta-6\sin^2\theta\cos\theta.\] Hence \[\frac{dy}{dx}=\cos^3\theta(1-6\sin^2\theta)=\cos^3\theta\bigl(1-6(1-\cos^2\theta)\bigr)=6\cos^5\theta-5\cos^3\theta.\]

(b) Since \(\sin\theta>0\), \(y=0\) implies \(1-2\sin^2\theta=0\), so \(\theta=\pi/4\). The point is \((1,0)\). Its tangent gradient is \[6\left(\frac1{\sqrt2}\right)^5-5\left(\frac1{\sqrt2}\right)^3=-\frac1{\sqrt2}.\] The normal gradient is therefore \(\sqrt2\), and \[y=\sqrt2(x-1)=\sqrt2x-\sqrt2.\]

For \(x>1/3\), the product and chain rules give \[\frac{dy}{dx}=-8e^{1-2x}\sqrt{3x-1}+\frac{6e^{1-2x}}{\sqrt{3x-1}}=\frac{e^{1-2x}}{\sqrt{3x-1}}\bigl(6-8(3x-1)\bigr).\] The prefactor is positive, so the derivative is zero when \[6-8(3x-1)=0\Rightarrow3x-1=\frac34\Rightarrow x=\frac7{12}.\] Substitution gives \[y=4e^{-1/6}\sqrt{\frac34}=2\sqrt3\,e^{-1/6}.\] The stationary point is \(\left(\frac7{12},2\sqrt3\,e^{-1/6}\right)\).

(a) Differentiate implicitly: \[10xy+5x^2\frac{dy}{dx}+8e^{2y}\frac{dy}{dx}-7=0.\] Thus \[\frac{dy}{dx}=\frac{7-10xy}{5x^2+8e^{2y}}.\] At \(y=0\), the curve equation gives \(4-7x+10=0\), so \(x=2\). Therefore \[\left.\frac{dy}{dx}\right|_{(2,0)}=\frac7{20+8}=\frac14.\]

(b) For every finite real \(x,y\), \(5x^2+8e^{2y}>0\). The denominator cannot be zero, so the gradient is finite everywhere on the curve. A tangent parallel to the \(y\)-axis is therefore impossible.

The product rule, followed by the chain rule, gives \[\frac{dy}{dx}=6\cos(x^2+1)+6x\bigl(-\sin(x^2+1)\bigr)(2x)=6\cos(x^2+1)-12x^2\sin(x^2+1).\]

(a) \[\frac{dx}{dt}=\frac{2(3t+4)-3(2t+1)}{(3t+4)^2}=\frac5{(3t+4)^2},\qquad\frac{dy}{dt}=\frac6{3t+4}.\] Therefore \[\frac{dy}{dx}=\frac{6/(3t+4)}{5/(3t+4)^2}=\frac65(3t+4),\qquad c=\frac65.\]

(b) From \(2\ln(3t+4)=\ln100\) and \(3t+4>0\), \[3t+4=10\Rightarrow t=2.\] Thus \[a=\frac{2(2)+1}{3(2)+4}=\frac12,\qquad m=\frac65(10)=12.\]

(c) The curve represents an increasing function, because \(3t+4>0\) and hence \(dy/dx>0\) throughout the domain.

For \(x\ne-3/2\), \[\frac{dy}{dx}=\frac{8(2x+3)-16x}{(2x+3)^2}-6=\frac{24}{(2x+3)^2}-6.\] Equating to zero gives \[(2x+3)^2=4\Rightarrow x=-\frac52\quad\text{or}\quad x=-\frac12.\] Substitution into the original curve gives the stationary points \[\left(-\frac52,30\right),\qquad\left(-\frac12,6\right).\]

(a) Substitution of \(y=e^{-1}\) gives \[-(x^2-3)+6x=14\Rightarrow x^2-6x+11=0.\] The discriminant is \(36-44=-8<0\), so there is no real \(x\), and no such point.

(b) Differentiate implicitly: \[2x\ln y+\frac{x^2-3}{y}\frac{dy}{dx}+6=0.\] Hence \[\frac{dy}{dx}=-\frac{y(2x\ln y+6)}{x^2-3}.\] At \((2,e^2)\), the gradient is \(-14e^2\). Therefore \[y-e^2=-14e^2(x-2)\Rightarrow y=-14e^2x+29e^2.\]

(a) The quotient rule gives \[\frac{dy}{dx}=\frac{4\cos x(3+\cos2x)+8\sin x\sin2x}{(3+\cos2x)^2}.\]

(b) Using \(\sin2x=2\sin x\cos x\) and \(\cos2x=2\cos^2x-1\), the numerator becomes \[4\cos x(2+2\cos^2x)+16\sin^2x\cos x=24\cos x-8\cos^3x=8\cos x(3-\cos^2x).\] The denominator is positive, and \(3-\cos^2x>0\). Thus stationary points require \(\cos x=0\), giving \(x=\pi/2,3\pi/2\). The coordinates are \[(\pi/2,2),\qquad(3\pi/2,-2).\]

(a) Differentiating implicitly, \[6e^{2x}y+3e^{2x}\frac{dy}{dx}+12e^{3x}+3y^2\frac{dy}{dx}=0.\] Divide by 3 and rearrange: \[(e^{2x}+y^2)\frac{dy}{dx}=-2e^{2x}y-4e^{3x},\] giving the required expression.

(b) Since \(e^{2x}+y^2>0\), a stationary point would require \[-2e^{2x}y-4e^{3x}=0\Rightarrow y=-2e^x.\] Substituting into the original equation gives \[-6e^{3x}+4e^{3x}-8e^{3x}=18\Rightarrow-10e^{3x}=18.\] This is impossible because \(e^{3x}>0\). Therefore the curve has no stationary points.

(a) \[\frac{dx}{dt}=\frac{2e^{2t}(e^{2t}+1)-2e^{2t}(e^{2t}-2)}{(e^{2t}+1)^2}=\frac{6e^{2t}}{(e^{2t}+1)^2},\quad\frac{dy}{dt}=3e^{3t}.\] Therefore \[\frac{dy}{dx}=\frac12e^t(e^{2t}+1)^2.\]

(b) At \(x=0\), \(e^{2t}=2\), so \(t=\frac12\ln2\), \(e^t=\sqrt2\). The gradient is \(\frac12\sqrt2(3)^2=9\sqrt2/2\).

Implicit differentiation, including the product rule, gives \[-6e^{-x}y^2+12e^{-x}y\frac{dy}{dx}+2e^{2x}-12\frac{dy}{dx}=0.\] At \((\ln3,2)\), \[-8+8y^\prime+18-12y^\prime=0,\] so \(10-4y^\prime=0\) and \(dy/dx=5/2\).

\[\frac{dy}{dx}=2\sec^2x-5\cos x.\] Setting this to zero gives \(2=5\cos^3x\), so \[x=\arccos\!\left(\sqrt[3]{2/5}\right)=0.742461890\ldots=0.742\text{ radians}.\]

Implicit differentiation gives \[2x\ln y+\frac{x^2}{y}\frac{dy}{dx}+2y\frac{dy}{dx}+4=0.\] At (2,1), \(0+4y^\prime+2y^\prime+4=0\), giving \(dy/dx=-2/3\).

\[\frac{dx}{dt}=-8\cos t\sin t=-4\sin2t,\quad\frac{dy}{dt}=2\sqrt3\cos2t.\] At \(t=\pi/6\), \((x,y)=(3,3/2)\) and \[m_{\rm tangent}=\frac{\sqrt3}{-2\sqrt3}=-\frac12,\quad m_{\rm normal}=2.\] Thus \(y-3/2=2(x-3)\), giving \(\boxed{4x-2y-9=0}\).

(a) \[\frac{dy}{dt}=\frac{\ln t-3}{t}+\frac{\ln t+2}{t}=\frac{2\ln t-1}{t},\quad\frac{dx}{dt}=\frac1{2\sqrt t}.\] Dividing gives \(dy/dx=(4\ln t-2)/\sqrt t\).

(b) The right-hand intercept B has \(t=e^3\). Its gradient is \((12-2)/e^{3/2}=10e^{-3/2}\).

(c) At M, \(4\ln t-2=0\), so \(t=e^{1/2}\). Hence \[M=\left(1+e^{1/4},\left(\frac12+2\right)\left(\frac12-3\right)\right)=\left(1+e^{1/4},-\frac{25}{4}\right).\] The quadratic in \(\ln t\) has its minimum at \(\ln t=1/2\).

Using the quotient rule, \[\frac{dy}{dx}=\frac{(1+2x)(3/x)-2(2+3\ln x)}{(1+2x)^2}.\] At \(x=1\), the gradient is \((9-4)/9=5/9\). Thus \[y-\frac23=\frac59(x-1),\qquad 5x-9y+1=0.\]

(a) \[\frac{dx}{dt}=8e^{2t},\qquad\frac{dy}{dt}=-5e^{-t}(\cos2t+2\sin2t).\] Therefore \[\frac{dy}{dx}=-\frac{5e^{-3t}}8(\cos2t+2\sin2t).\]

(b) At a stationary point, \(\cos2t+2\sin2t=0\), so \(\tan2t=-1/2\). The unique value in the given interval is \(t=-\frac12\tan^{-1}(1/2)=-0.2318238\ldots\). The derivative changes from positive to negative there, so this is the maximum. Substitution gives \(M=(2.52,5.64)\), to 3 significant figures.

(a) The intercepts satisfy \((x-1)(x-4)=0\), so \(B\) has \(x=4\). By the product rule, \[\frac{dy}{dx}=e^{-x/2}\left(2x-5-\frac12(x^2-5x+4)\right)=-\frac12e^{-x/2}(x-2)(x-7).\] At \(x=4\), this is \(3e^{-2}\).

(b) The stationary points have \(x=2,7\). The derivative changes from positive to negative at \(x=7\), so \(C\) is the maximum. Its ordinate is \(e^{-7/2}(49-35+4)=18e^{-7/2}\), giving \(C=(7,18e^{-7/2})\).

(a) \[\frac{dx}{dt}=\frac1{(t+2)^2},\qquad\frac{dy}{dt}=2t+a,\qquad\frac{dy}{dx}=(2t+a)(t+2)^2,\quad t\ne-2.\] Set this equal to 1 and expand to obtain the stated cubic.

(b) Substituting \(t=-1\) gives \(-2+(a+8)-(4a+8)+4a-1=0\), so \(a=3\).

(c) With \(a=3\), the cubic factors as \[(t+1)(2t^2+9t+11)=0.\] The quadratic has discriminant \(9^2-4(2)(11)=-7<0\). Thus the only real parameter is \(t=-1\), giving the unique point \(P=(1,-1)\).

(a) The quotient rule gives \[\frac{dy}{dx}=\frac{8e^{2x}(e^x+2)-(4e^{2x}+9)e^x}{(e^x+2)^2}=\frac{e^x(4e^{2x}+16e^x-9)}{(e^x+2)^2}.\] At \(P\), \(x=0\), so the gradient is \(11/9\).

(b) At a stationary point, \(4e^{2x}+16e^x-9=0\), or \((2e^x-1)(2e^x+9)=0\). Because \(e^x>0\), \(e^x=1/2\), giving \(x=-\ln2\). Substitution gives \(y=(1+9)/(1/2+2)=4\). The derivative changes from negative to positive, confirming \(M=(-\ln2,4)\) is the minimum.

(a) At \(P\), \(3\sin2t-4\sin t=2\sin t(3\cos t-2)=0\). Since \(0<t<\pi/2\), \(\sin t\ne0\), so \(\cos t=2/3\).

(b) \[\frac{dx}{dt}=k\sec^2t,\qquad\frac{dy}{dt}=6\cos2t-4\cos t.\] Thus \[\frac{dy}{dx}=\frac{(6(2\cos^2t-1)-4\cos t)\cos^2t}{k}=\frac{12\cos^4t-4\cos^3t-6\cos^2t}{k}.\]

(c) At \(\cos t=2/3\), the tangent gradient is \(-40/(27k)\). A normal gradient \(9/10\) requires tangent gradient \(-10/9\). Therefore \(-40/(27k)=-10/9\), giving \(k=4/3\).

Differentiate implicitly, using the product rule: \[8e^{2x}y+4e^{2x}\frac{dy}{dx}+2y\frac{dy}{dx}=0.\] At \((0,-7)\), \[-56+(4-14)\frac{dy}{dx}=0.\] Therefore \[\frac{dy}{dx}=-\frac{28}{5}.\]

(a) A has \(\ln x=0\), so \(x=1\). The quotient rule gives \[\frac{dy}{dx}=\frac{2(3x+1)/x-6\ln x}{(3x+1)^2}.\] At \(x=1\), the gradient is \(8/16=1/2\).

(b) A stationary point has \(N(x)=2(3x+1)/x-6\ln x=0\). The numerator is continuous for \(x>0\), and \[N(3.0)\approx0.07499>0,\qquad N(3.1)\approx-0.14325<0.\] Therefore the stationary point B lies between 3.0 and 3.1. Also \(N\) decreases strictly, so the derivative changes from positive to negative.

(c) Here \(h=(3-1)/2=1\), with ordinates \[y_0=0,\quad y_1=\frac{2\ln2}{7},\quad y_2=\frac{2\ln3}{10}.\] Thus \[A\approx\frac12\left[0+2\left(\frac{2\ln2}{7}\right)+\frac{2\ln3}{10}\right]\approx0.31.\]

(a) \[\frac{dx}{d\theta}=-6\sin2\theta,\qquad\frac{dy}{d\theta}=4\cos\theta.\] In the interior of the given interval, \[\frac{dy}{dx}=\frac{4\cos\theta}{-6\sin2\theta}=-\frac1{3\sin\theta}.\] At P the gradient is 2, so \(\sin\theta=-1/6\). Only the quadrant III value is in the interval: \[\theta=\pi+\sin^{-1}(1/6)\approx3.31\text{ radians}.\]

(b) Substituting the parametric equations into the line, \[9\cos2\theta+4\sin\theta=0.\] Let \(u=\sin\theta\), and use \(\cos2\theta=1-2u^2\): \[18u^2-4u-9=0,\qquad u=\frac{2\pm\sqrt{166}}{18}.\] Since \(-1\le\sin\theta\le0\) on the given interval, \[\sin\theta=\frac{2-\sqrt{166}}{18}\approx-0.604672.\] Hence the gradient at Q is \[\frac{dy}{dx}=-\frac1{3\sin\theta}=\frac6{\sqrt{166}-2}\approx0.551.\]

Differentiate implicitly using the product rule: \[2xy+x^2\frac{dy}{dx}+6y^2\frac{dy}{dx}=0.\] Thus \[\frac{dy}{dx}=-\frac{2xy}{x^2+6y^2}.\] At \((4,2)\), the tangent gradient is \(-16/40=-2/5\), so the normal gradient is \(5/2\). Therefore \[y-2=\frac52(x-4),\] or \[\boxed{5x-2y-16=0}.\]

(a) By the quotient rule, \[\frac{dy}{dx}=\frac{18e^{2x}(e^x-1)-e^x(9e^{2x}+16)}{(e^x-1)^2}.\] Factorising the numerator gives \[\frac{dy}{dx}=\frac{e^x(9e^{2x}-18e^x-16)}{(e^x-1)^2}=\frac{e^x(3e^x-8)(3e^x+2)}{(e^x-1)^2}.\] Setting the derivative to zero gives the required equation.

(b) Since \(e^x>0\), neither \(e^x\) nor \(3e^x+2\) can be zero. The only possibility is \[e^x=8/3,\qquad x=\ln(8/3).\] It is in the curve domain because \(e^x\ne1\). Substitution gives \[y=\frac{9(8/3)^2+16}{8/3-1}=48.\] The unique stationary point is \[\boxed{\bigl(\ln(8/3),48\bigr)}.\]

Write \(y=x^{-2}\ln x\), and differentiate by the product rule: \[\frac{dy}{dx}=-2x^{-3}\ln x+x^{-3}=\frac{1-2\ln x}{x^3}.\] At \(x=e\), \(\ln e=1\), so \[\boxed{\frac{dy}{dx}=-\frac1{e^3}}.\]

Differentiate implicitly, applying both the product rule and chain rule: \[2e^{2x}\cos2y-2e^{2x}\sin2y\frac{dy}{dx}+\cos y\frac{dy}{dx}=0.\] At \((0,\pi/6)\), \[1+\left(-\sqrt3+\frac{\sqrt3}2\right)\frac{dy}{dx}=0.\] Therefore \[\boxed{\frac{dy}{dx}=\frac2{\sqrt3}=\frac{2\sqrt3}3}.\]

By the chain rule, \[\frac{dy}{dx}=4\frac2{2x+5}=\frac8{2x+5}.\] At \(x=-2\), the tangent gradient is 8. Thus \[y-7=8(x+2),\] giving \[\boxed{y=8x+23}.\]

(a) Differentiate with respect to t: \[\frac{dx}{dt}=\frac2{2t+3},\qquad\frac{dy}{dt}=\frac{2(2t+3)-2(2t-3)}{(2t+3)^2}=\frac{12}{(2t+3)^2}.\] Divide the two derivatives to obtain \[\boxed{\frac{dy}{dx}=\frac{12}{(2t+3)^2}\div\frac2{2t+3}=\frac6{2t+3}}.\]

(b) At A, \(x=0\), so \(\ln(2t+3)=0\). Therefore \(2t+3=1\) and \(t=-1\). The gradient is \[\boxed{6}.\]

(c) At B, \(y=0\), so \(2t-3=0\), giving \(t=3/2\). Thus the gradient is \[\boxed{6/(2(3/2)+3)=1}.\]

Differentiate: \[\frac{dy}{dx}=5-4\sec^2 2x.\] At a stationary point, \[\sec^2 2x=5/4,\qquad\cos^2 2x=4/5.\] Since \(0\le2x<\pi/2\), the cosine is positive, so \[x=\frac12\arccos\frac2{\sqrt5}=0.231823\ldots.\] Here \(\tan2x=1/2\), giving \[y=5x-1=0.159119\ldots.\] Thus the coordinates are \[\boxed{(0.232,0.159)}.\]

Differentiate implicitly, using the product and chain rules: \[2x+4\cos3y-12x\sin3y\frac{dy}{dx}=0.\] Hence \[\frac{dy}{dx}=\frac{2x+4\cos3y}{12x\sin3y}.\] At the given point, \(3y=\pi/4\) and both sine and cosine equal \(1/\sqrt2\). Thus \[m_{\rm tangent}=\frac{2\sqrt2+4/\sqrt2}{12\sqrt2/\sqrt2}=\frac{4\sqrt2}{12}=\frac{\sqrt2}{3}.\] The normal has the negative reciprocal gradient: \[\boxed{m_{\rm normal}=-\frac3{\sqrt2}=-\frac{3\sqrt2}{2}}.\]

(a) At the intercepts, \[\ln x(\ln x-2)=0,\qquad x=1\text{ or }e^2.\] Thus A has x-coordinate 1 and B has x-coordinate \(e^2\). Differentiate using the chain rule: \[\frac{dy}{dx}=\frac{2\ln x}{x}-\frac2x=\frac{2\ln x-2}{x}.\] Substituting gives \[\boxed{m_A=-2,\qquad m_B=\frac2{e^2}=2e^{-2}}.\]

(b) At M, \(2\ln x-2=0\), so \(\ln x=1\), giving \[\boxed{x_M=e}.\] The derivative changes from negative to positive there, confirming a minimum.

(a) The logarithms require \(t>0\). Combine the logarithms and exponentiate: \[\ln\frac{2t+6}{t}=\ln4,\qquad2t+6=4t,\qquad\boxed{t=3}.\]

(b) Differentiate with respect to t: \[\frac{dx}{dt}=\frac2{2t+6}-\frac1t,\qquad\frac{dy}{dt}=\ln t+1.\] At t=3, \(dx/dt=1/6-1/3=-1/6\) and \(dy/dt=\ln3+1\). Therefore \[\boxed{\left.\frac{dy}{dx}\right|_P=\frac{\ln3+1}{-1/6}=-6(\ln3+1)}.\]

Apply the product and chain rules: \[\frac{dx}{dt}=2e^{2t}\cos4t-4e^{2t}\sin4t,\qquad\frac{dy}{dt}=6\cos2t.\] At t=0 these derivatives are 2 and 6 respectively. Therefore \[\boxed{\left.\frac{dy}{dx}\right|_{t=0}=\frac6{2}=3}.\]