(i) sin x + cos y = xy, find dy/dx. (2)
(ii) x = a cos³t, y = a sin³t, find dy/dx. (2)
(iii) If y = (sin⁻¹x)², then show that (1 − x²)(d²y/dx²) − x(dy/dx) = 2. (2)

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Accepted Answer

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> [!TLDR]
> (i) Differentiate implicitly, collecting all $y'$ terms on one side. (ii) Use the parametric-derivative formula $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$. (iii) Differentiate $y=(\sin^{-1}x)^2$ twice and combine to eliminate the $\sqrt{1-x^2}$.
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> **(i) $\sin x+\cos y=xy$, find $\dfrac{dy}{dx}$.**
>
> Differentiate both sides w.r.t. $x$ (using the product rule on $xy$, and chain rule on $\cos y$):
> $$\cos x-\sin y\cdot y'=y+x y'$$
> Collect $y'$ terms:
> $$\cos x-y=xy'+\sin y\cdot y'=y'(x+\sin y)$$
> $$y'=\dfrac{\cos x-y}{x+\sin y}$$
>
> **(ii) $x=a\cos^3 t$, $y=a\sin^3 t$, find $\dfrac{dy}{dx}$.**
>
> $$\dfrac{dx}{dt}=3a\cos^2t\cdot(-\sin t)=-3a\cos^2t\sin t$$
> $$\dfrac{dy}{dt}=3a\sin^2t\cdot\cos t$$
> $$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3a\sin^2t\cos t}{-3a\cos^2t\sin t}=-\dfrac{\sin t}{\cos t}=-	an t$$
>
> **(iii) $y=(\sin^{-1}x)^2$, show $(1-x^2)y''-xy'=2$.**
>
> Differentiate once (chain rule):
> $$y'=2\sin^{-1}x\cdot\dfrac{1}{\sqrt{1-x^2}}=\dfrac{2\sin^{-1}x}{\sqrt{1-x^2}}$$
> …

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

Q.(i) sin x + cos y = xy, find dy/dx.

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Q.(i) sin x + cos y = xy, find dy/dx.

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