Implicit Differentiation

The point is on the curve: $25+144=169$.

$2x+2y\frac{dy}{dx}=0\ \Longrightarrow\ \frac{dy}{dx}=-\frac{x}{y}\quad(y\neq0)$

At $(5,-12)$ the slope is $-\frac{5}{-12}=\frac{5}{12}$, so $y+12=\frac{5}{12}(x-5)$, that is $5x-12y=169$.

Check: the distance from the origin to that line is $\frac{169}{\sqrt{169}}=13$, the radius — exactly what a tangent to this circle must satisfy.

Check the point: $2+2=4$. Differentiating $x^{1/2}+y^{1/2}=4$,

$\tfrac12x^{-1/2}+\tfrac12y^{-1/2}\frac{dy}{dx}=0\ \Longrightarrow\ \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}$

At $(4,4)$ the slope is $-1$, so $y-4=-(x-4)$, i.e. $x+y=8$.

The curve is $y=\left(4-\sqrt{x}\right)^{2}$ for $0\le x\le16$, but the formula needs $x>0$ and $y>0$: valid on $0<x<16$. The tangent is vertical at $(0,16)$ and horizontal at $(16,0)$.

The point is on the curve: $2+4=6$. Differentiating term by term,

$2xy+x^{2}\frac{dy}{dx}+y^{2}+2xy\frac{dy}{dx}=0$

Collect and factor: $\frac{dy}{dx}\left(x^{2}+2xy\right)=-\left(2xy+y^{2}\right)$, so

$\frac{dy}{dx}=-\frac{2xy+y^{2}}{x^{2}+2xy}=-\frac{y(2x+y)}{x(x+2y)}$

At $(1,2)$: $-\dfrac{2(2+2)}{1(1+4)}=-\dfrac{8}{5}$.

$\cos(x+y)\left(1+\frac{dy}{dx}\right)=2-2\frac{dy}{dx}$

Collect: $\frac{dy}{dx}\left(2+\cos(x+y)\right)=2-\cos(x+y)$, giving the formula above.

At $(\pi,\pi)$, $\cos(2\pi)=1$, so the slope is $\frac{2-1}{2+1}=\frac13$ and $y-\pi=\frac13(x-\pi)$, i.e. $y=\frac{x}{3}+\frac{2\pi}{3}$.

A vertical tangent needs $2+\cos(x+y)=0$, but $1\le2+\cos(x+y)\le3$. The denominator never vanishes, so $\frac{dy}{dx}$ exists at every point of the curve.

$3x^{2}+3y^{2}\frac{dy}{dx}=0$ gives $\frac{dy}{dx}=-\frac{x^{2}}{y^{2}}$. Differentiate again, remembering $\frac{d}{dx}y^{2}=2y\frac{dy}{dx}$:

$\frac{d^2y}{dx^2}=-\frac{2xy^{2}-x^{2}\left(2y\frac{dy}{dx}\right)}{y^{4}}=-\frac{2x\left(y-x\frac{dy}{dx}\right)}{y^{3}}$

Substituting $\frac{dy}{dx}=-\frac{x^{2}}{y^{2}}$: $y-x\frac{dy}{dx}=y+\frac{x^{3}}{y^{2}}=\frac{x^{3}+y^{3}}{y^{2}}=\frac{1}{y^{2}}$, using the original equation.

Hence $\frac{d^2y}{dx^2}=-\frac{2x}{y^{3}}\cdot\frac{1}{y^{2}}=-\frac{2x}{y^{5}}$.

The point is on the curve: $2(10)^{2}=200=25(9-1)$. Differentiating, with the chain rule on the outer square,

$4\left(x^{2}+y^{2}\right)\left(2x+2y\frac{dy}{dx}\right)=25\left(2x-2y\frac{dy}{dx}\right)$

Halve both sides and substitute $x=3$, $y=1$, $x^{2}+y^{2}=10$:

$40\left(3+\frac{dy}{dx}\right)=25\left(3-\frac{dy}{dx}\right)$

So $65\frac{dy}{dx}=-45$ and $\frac{dy}{dx}=-\frac{9}{13}$. Then $y-1=-\frac{9}{13}(x-3)$, i.e. $9x+13y=40$.