a) Find (dy)/(dx) given i x=ln(2t^2) and y=ln(4+t^2) (3 Marks) ii x^3+3x^2y^3-4xy^2=204 at (5,-1) (3 Marks)

a) i) To find $\frac{dy}{dx}$, we need to use the chain rule. Let's start by finding $\frac{dx}{dt}$ and $\frac{dy}{dt}$.<br /><br />Given $x = \ln(2t^2)$, we can differentiate with respect to $t$ to get:<br />$\frac{dx}{dt} = \frac{1}{t} \cdot 2t = 2$<br /><br />Given $y = \ln(4 + t^2)$, we can differentiate with respect to $t$ to get:<br />$\frac{dy}{dt} = \frac{1}{4 + t^2} \cdot 2t = \frac{2t}{4 + t^2}$<br /><br />Now, we can use the chain rule to find $\frac{dy}{dx}$:<br />$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2t}{4 + t^2}}{2} = \frac{t}{4 + t^2}$<br /><br />ii) To find $\frac{dy}{dx}$ at the point $(5, -1)$, we need to use implicit differentiation.<br /><br />Given $x^3 + 3x^2y^3 - 4xy^2 = 204$, we can differentiate both sides with respect to $x$:<br />$3x^2 + 6xy^2 \frac{dy}{dx} + 3x^2y^3 \frac{dy}{dx} - 4y^2 - 8xy \frac{dy}{dx} = 0$<br /><br />Now, we can substitute the point $(5, -1)$ into the equation:<br />$3(5)^2 + 6(5)(-1)^2 \frac{dy}{dx} + 3(5)^2(-1)^3 \frac{dy}{dx} - 4(-1)^2 - 8(5)(-1) \frac{dy}{dx} = 0$<br /><br />Simplifying the equation:<br />$75 + 30 \frac{dy}{dx} - 75 \frac{dy}{dx} + 4 + 40 \frac{dy}{dx} = 0$<br /><br />Combining like terms:<br />$-5 \frac{dy}{dx} = -79$<br /><br />Solving for $\frac{dy}{dx}$:<br />$\frac{dy}{dx} = \frac{79}{5}$<br /><br />Therefore, the answer is:<br />i) $\frac{dy}{dx} = \frac{t}{4 + t^2}$<br />ii) $\frac{dy}{dx} = \frac{79}{5}$ at $(5, -1)$