Here are the solutions for both parts of question 10. 10. (i) Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ if $y = \ln\left(\frac{1-x}{1+x}\right)$ Given the function: $$y = \ln\left(\frac{1-x}{1+x}\right)$$ We can use the logarithm property $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$ to simplify the expression: $$y = \ln(1-x) - \ln(1+x)$$ Step 1: Find $\frac{dy}{dx}$. Differentiate $y$ with respect to $x$. Recall that $\frac{d}{dx}(\ln(u)) = \frac{1}{u} \frac{du}{dx}$. $$\frac{dy}{dx} = \frac{1}{1-x} \cdot \frac{d}{dx}(1-x) - \frac{1}{1+x} \cdot \frac{d}{dx}(1+x)$$ $$\frac{dy}{dx} = \frac{1}{1-x} \cdot (-1) - \frac{1}{1+x} \cdot (1)$$ $$\frac{dy}{dx} = -\frac{1}{1-x} - \frac{1}{1+x}$$ To combine the terms, find a common denominator: $$\frac{dy}{dx} = \frac{-(1+x) - (1-x)}{(1-x)(1+x)}$$ $$\frac{dy}{dx} = \frac{-1-x-1+x}{1-x^2}$$ $$\frac{dy}{dx} = \frac{-2}{1-x^2}$$ This can also be written as: $$\frac{dy}{dx} = \frac{2}{x^2-1}$$ Step 2: Find $\frac{d^2y}{dx^2}$. Now, differentiate $\frac{dy}{dx}$ with respect to $x$. We can rewrite $\frac{dy}{dx}$ as $2(x^2-1)^{-1}$. $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(2(x^2-1)^{-1}\right)$$ Using the chain rule, $\frac{d}{dx}(u^n) = nu^{n-1}\frac{du}{dx}$: $$\frac{d^2y}{dx^2} = 2 \cdot (-1)(x^2-1)^{-2} \cdot \frac{d}{dx}(x^2-1)$$ $$\frac{d^2y}{dx^2} = -2(x^2-1)^{-2} \cdot (2x)$$ $$\frac{d^2y}{dx^2} = -\frac{4x}{(x^2-1)^2}$$ Note: The instruction "giving your answer in terms of t" appears to be a typo, as there is no variable 't' in the given function. The answers are provided in terms of x. The derivatives are: $$\boxed{\frac{dy}{dx} = \frac{2}{x^2-1}}$$ $$\boxed{\frac{d^2y}{dx^2} = -\frac{4x}{(x^2-1)^2}}$$ 10. (ii) If $x^2 - xy + y^2 = 3$, show that $(2y-x)\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 - 2\frac{dy}{dx} + 2 = 0$ Given the implicit equation: $$x^2 - xy + y^2 = 3$$ Step 1: Find $\frac{dy}{dx}$ by implicitly differentiating the equation with respect to $x$. Differentiate each term: $$\frac{d}{dx}(x^2) - \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(3)$$ $$2x - \left(1 \cdot y + x \cdot \frac{dy}{dx}\right) + 2y \frac{dy}{dx} = 0$$ $$2x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$$ Group terms containing $\frac{dy}{dx}$: $$(2y-x)\frac{dy}{dx} = y - 2x$$ Step 2: Find $\frac{d^2y}{dx^2}$ by implicitly differentiating the equation from Step 1 with respect to $x$. We will differentiate $(2y-x)\frac{dy}{dx} = y - 2x$. Apply the product rule to the left side and differentiate the right side: $$\frac{d}{dx}(2y-x) \cdot \frac{dy}{dx} + (2y-x) \cdot \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(y - 2x)$$ $$\left(2\frac{dy}{dx} - 1\right)\frac{dy}{dx} + (2y-x)\frac{d^2y}{dx^2} = \frac{dy}{dx} - 2$$ Expand the left side: $$2\left(\frac{dy}{dx}\right)^2 - \frac{dy}{dx} + (2y-x)\frac{d^2y}{dx^2} = \frac{dy}{dx} - 2$$ Rearrange the terms to match the target equation: $$(2y-x)\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 - \frac{dy}{dx} - \frac{dy}{dx} + 2 = 0$$ $$(2y-x)\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 - 2\frac{dy}{dx} + 2 = 0$$ This matches the equation we needed to show.