Implicit Differentiation

$2x + 3(y + xy') - 3y^2y' = 0$.

$3xy$ used the product rule ($y + xy'$); $y^3$ became $3y^2y'$.

$3xy' - 3y^2y' = -2x - 3y$.

$y'(3x - 3y^2) = -2x - 3y$, so $y' = \dfrac{-2x - 3y}{3x - 3y^2} = \dfrac{2x + 3y}{3y^2 - 3x}$.

Product rule on the left: $7^{y} + x \cdot 7^{y}\log 7\,y' = 2y\,y'$.

$7^{y}$ is a constant base to a $y$ power: $(7^{y})' = 7^{y}\log 7\,y'$. $y^2$ became $2yy'$.

$x\,7^{y}\log 7\,y' - 2y\,y' = -7^{y}$.

$y'(x\,7^{y}\log 7 - 2y) = -7^{y}$, so $y' = \dfrac{7^{y}}{2y - x\,7^{y}\log 7}$.

$\cos(x + y)(1 + y') = y'\cos x - y\sin x$.

Chain rule on the left: the inside $x + y$ differentiates to $1 + y'$. Product rule on the right.

$\cos(x + y)\,y' - \cos x\,y' = -y\sin x - \cos(x + y)$.

$y'\big(\cos(x + y) - \cos x\big) = -\big(y\sin x + \cos(x + y)\big)$, so $y' = \dfrac{y\sin x + \cos(x + y)}{\cos x - \cos(x + y)}$.

$e^{xy}(y + xy') = 1 + 2y\,y'$.

Chain rule on $e^{xy}$, with the product rule on the inside $xy$.

$xe^{xy}\,y' - 2y\,y' = 1 - ye^{xy}$.

$y'(xe^{xy} - 2y) = 1 - ye^{xy}$, so $y' = \dfrac{1 - ye^{xy}}{xe^{xy} - 2y}$.

$1 + 2 + 8 = 11$. On the curve.

$2x + y + xy' + 3y^2y' = 0$.

At $(1, 2)$: $2 + 2 + y' + 12y' = 0 \Rightarrow 13y' = -4 \Rightarrow m = -\dfrac{4}{13}$.

$y - 2 = -\dfrac{4}{13}(x - 1)$.

$1 + \sin\pi = 1 + 0 = 1 = 1^2$. On the curve.

$y' + \cos(\pi xy)\cdot\pi(y + xy') = 2x$.

At $(1, 1)$, $\cos\pi = -1$: $y' - \pi(1 + y') = 2 \Rightarrow y'(1 - \pi) = 2 + \pi \Rightarrow m = \dfrac{2 + \pi}{1 - \pi}$.

$y - 1 = \dfrac{2 + \pi}{1 - \pi}(x - 1)$.

$2e^{0} + 0 = 2$. On the curve.

$y'e^{x} + ye^{x} + 2xy^2 + 2x^2y\,y' = 0$.

At $(0, 2)$: $y' + 2 + 0 + 0 = 0 \Rightarrow m = -2$.

$y - 2 = -2(x - 0)$.

$2x + y + xy' + 2yy' = 0 \Rightarrow y' = -\dfrac{2x + y}{x + 2y}$. So $N = 2x + y$, $D = x + 2y$.

$N = 0$: $y = -2x$. Then $x^2 - 2x^2 + 4x^2 = 3x^2 = 12$, so $x = \pm 2$: points $(2, -4)$ and $(-2, 4)$.

$D = 0$: $x = -2y$. Then $4y^2 - 2y^2 + y^2 = 3y^2 = 12$, so $y = \pm 2$: points $(-4, 2)$ and $(4, -2)$.

$D$ at $(2, -4)$ is $-6$, at $(-2, 4)$ is $6$. $N$ at $(-4, 2)$ is $-6$, at $(4, -2)$ is $6$. None are $0$.

Horizontal: $y = -4$ and $y = 4$. Vertical: $x = -4$ and $x = 4$.

$2yy' - 2y' = 3x^2 - 3 \Rightarrow y' = \dfrac{3x^2 - 3}{2y - 2}$.

$N = 0$: $x = \pm 1$. At $x = 1$: $y^2 - 2y + 3 = 0$, discriminant $4 - 12 < 0$, no point. At $x = -1$: $y^2 - 2y - 1 = 0$, so $y = 1 \pm \sqrt{2}$.

$D = 0$: $y = 1$. Then $-1 = x^3 - 3x - 1$, so $x(x^2 - 3) = 0$: $x = 0$, $\pm\sqrt{3}$.

At $y = 1 \pm \sqrt{2}$, $D = \pm 2\sqrt{2} \neq 0$. At $x = 0$, $N = -3$; at $x = \pm\sqrt{3}$, $N = 6$. None are $0$.

Horizontal: $y = 1 + \sqrt{2}$ and $y = 1 - \sqrt{2}$. Vertical: $x = 0$, $x = \sqrt{3}$, $x = -\sqrt{3}$.

$2x + 8yy' = 0 \Rightarrow y' = -\dfrac{x}{4y}$.

$y'' = -\dfrac{(1)(4y) - x(4y')}{16y^2} = -\dfrac{y - xy'}{4y^2}$.

$y - x\left(-\dfrac{x}{4y}\right) = \dfrac{4y^2 + x^2}{4y}$, so $y'' = -\dfrac{x^2 + 4y^2}{16y^3}$.

On the curve $x^2 + 4y^2 = 16$: $y'' = -\dfrac{16}{16y^3} = -\dfrac{1}{y^3}$.

$3x^2 + 3y^2y' = 0 \Rightarrow y' = -\dfrac{x^2}{y^2}$. At $(1, 2)$: $y' = -\dfrac{1}{4}$.

$y'' = -\dfrac{2x\,y^2 - x^2 \cdot 2y\,y'}{y^4}$.

At $(1, 2)$ with $y' = -\tfrac{1}{4}$: $y'' = -\dfrac{2 \cdot 4 - 1 \cdot 4 \cdot \left(-\frac{1}{4}\right)}{16} = -\dfrac{8 + 1}{16}$.

$y'' = -\dfrac{9}{16}$. (In general $y'' = -\dfrac{2x(x^3 + y^3)}{y^5} = -\dfrac{18x}{y^5}$, which agrees.)

$\cos y\,y' = 1 \Rightarrow y' = \dfrac{1}{\cos y}$.

$y' = (\cos y)^{-1}$, so $y'' = -(\cos y)^{-2}(-\sin y)\,y' = \dfrac{\sin y}{\cos^2 y}\,y'$.

$y'' = \dfrac{\sin y}{\cos^2 y}\cdot\dfrac{1}{\cos y} = \dfrac{\sin y}{\cos^3 y}$.

On the curve $\sin y = x$: $y'' = \dfrac{x}{\cos^3 y}$.