음함수의 미분법

$\frac{dy}{dx}=\frac{-3{x}^{2}y-{y}^{3}}{{x}^{3}+3x{y}^{2}}$

Remember that $y$ is a function of $x$. We begin with implicit differentiation.

$$\begin{aligned} xy + e^x + e^y &= 1\\ y+x\frac{\mathrm{d}y}{\mathrm{d}x}+e^x+e^y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{Now, we solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$.} \\ x\frac{\mathrm{d}y}{\mathrm{d}x}+e^y\frac{\mathrm{d}y}{\mathrm{d}x}&=-(e^x+y)\\ (x+e^y)\frac{\mathrm{d}y}{\mathrm{d}x}&=-(e^x+y)\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{e^x+y}{e^y+x} \end{aligned}$$

Differentiate both sides of the equation with respect to $x$:

$$\begin{aligned} e^y\frac{\mathrm{d}y}{\mathrm{d}x}&=x\cdot2y\frac{\mathrm{d}y}{\mathrm{d}x}+y^2+1 \\ \text{Now, get the derivative on one side and solve} \\ e^y\frac{\mathrm{d}y}{\mathrm{d}x}-2xy\frac{\mathrm{d}y}{\mathrm{d}x}&=y^2+1\\ \frac{\mathrm{d}y}{\mathrm{d}x}\left(e^y-2xy\right)&=y^2+1\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{y^2+1}{e^y-2xy} \end{aligned}$$

• First we find the $x$-coordinates where $y=1$.

$$\begin{aligned} x^2\tan\left(\frac{\pi}{4}\right)+2x\log(1) &= 16 \\ x^2\cdot 1 +2x\cdot 0 &=16\\ x^2 &= 16\\ \end{aligned}$$

So $x=\pm 4$.

• Now we use implicit differentiation to get $y'$ in terms of $x,y$:

$$\begin{aligned} x^2\tan(\pi y/4)+2x\log(y) &= 16\\ 2x\tan(\pi y/4) + x^2 \frac{\pi}{4}\sec^2(\pi y/4)\cdot y' + 2\log(y) + \frac{2x}{y} \cdot y' &= 0\,.\\ \end{aligned}$$

• Now set $y=1$ and use $\tan(\pi/4)=1\,, \sec(\pi/4)=\sqrt{2}$ to get

$$\begin{aligned} 2x\tan(\pi/4) + x^2 \frac{\pi}{4} \sec^2(\pi/4)y' + 2\log(1) + 2x\cdot y' &= 0\\ 2x + \frac{\pi}{2} x^2 y' +2x y' &= 0 \\ y' = -\frac{2x}{x^2 \pi/2 + 2x} &= -\frac{4}{\pi x + 4} \\ \end{aligned}$$

• So at $(x,y)=(4,1)$ we have $y' = -\dfrac{4}{4\pi+4} = -\dfrac{1}{\pi + 1}$
• and at $(x,y)=(-4,1)$ we have $y' = \dfrac{1}{\pi-1}$

Differentiate the equation and solve:

$$\begin{aligned} 3x^2 + 4y^3 \frac{\mathrm{d}y}{\mathrm{d}x} &= -\sin(x^2+y) \cdot\left(2x + \frac{\mathrm{d}y}{\mathrm{d}x}\right) \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= -\frac{2x\sin(x^2+y)+3x^2}{4y^3+\sin(x^2+y)} \end{aligned}$$

• First we find the $x$-coordinates where $y=0$.

$$\begin{aligned} x^2e^0+4x\cos(0) &= 5 \\ x^2 +4x - 5 &=0\\ (x+5)(x-1)&=0 \end{aligned}$$

So $x=1,-5$.

• Now we use implicit differentiation to get $y'$ in terms of $x,y$:

$$\begin{aligned} x^2e^y+4x\cos(y) &= 5 & \text{differentiate both sides} \\ x^2 \cdot e^y \cdot y' + 2x e^y + 4x(-\sin(y)) \cdot y' + 4\cos(y) &= 0 \end{aligned}$$

• Now set $y=0$ to get

$$\begin{aligned} x^2 \cdot e^0 \cdot y' + 2x e^0 + 4x(-\sin(0)) \cdot y' + 4\cos(0) &= 0 \\ x^2y' + 2x + 4 &=0 \\ y' &= - \frac{4+2x}{x^2}. \end{aligned}$$

• So at $(x,y)=(1,0)$ we have $y' = -6$,
• and at $(x,y)=(-5,0)$ we have $y' = \frac{6}{25}$.

Differentiate the equation and solve:

$$\begin{aligned} 2x + 2y \frac{\mathrm{d}y}{\mathrm{d}x} &= \cos(x+y) \cdot\left(1 + \frac{\mathrm{d}y}{\mathrm{d}x}\right) \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\cos(x+y)-2x}{2y-\cos(x+y)} \end{aligned}$$

• First we find the $x$-coordinates where $y=0$.

$$\begin{aligned} x^2\cos(0)+2xe^0 &= 8 \\ x^2 +2x - 8 &=0\\ (x+4)(x-2)&=0 \end{aligned}$$

So $x=2,-4$.

• Now we use implicit differentiation to get $y'$ in terms of $x,y$:

$$\begin{aligned} x^2\cos(y)+2xe^y &= 8 & \text{differentiate both sides} \\ x^2 \cdot (-\sin y) \cdot y' + 2x \cos y + 2xe^y \cdot y' + 2e^y &= 0 \end{aligned}$$

• Now set $y=0$ to get

$$\begin{aligned} x^2 \cdot (-\sin 0) \cdot y' + 2x \cos 0 + 2xe^0 \cdot y' + 2e^0 &= 0 \\ 0 + 2x + 2xy' + 2 &=0 \\ y' &= - \frac{2+2x}{2x} = -\frac{1+x}{x} \end{aligned}$$

• So at $(x,y)=(2,0)$ we have $y' = -\frac{3}{2}$,
• and at $(x,y)=(-4,0)$ we have $y' = -\frac{3}{4}$.

The question asks at which points on the ellipse $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}{}=1$. So, we begin by differentiating, implicitly:

$$\begin{aligned} 2x+6y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{We could solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ at this point, but it's not necessary. We want to know when $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ is equal to one:} \\ 2x+6y(1)&=0\\ x&=-3y \\ \text{That is, $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=1$ at those points along the ellipse where $x=-3y$. We plug this into the equation of the ellipse to find the coordinates of these points.} \\ \left(-3y\right)^2+3y^2&=1\\ 12y^2&=1\\ y=\pm\frac{1}{\sqrt{12}}=\pm\frac{1}{2\sqrt{3}} \end{aligned}$$

So, the points along the ellipse where the tangent line is parallel to the line $y=x$ occur when $y=\dfrac{1}{2\sqrt{3}}$ and $x=-3y$, and when $y=\dfrac{-1}{2\sqrt{3}}$ and $x=-3y$. That is, the points $\left(\dfrac{-\sqrt{3}}{2},\dfrac{1}{2\sqrt{3}}\right)$ and $\left(\dfrac{\sqrt{3}}{2},\dfrac{-1}{2\sqrt{3}}\right)$.

First, we differentiate implicitly with respect to $x$.

$$\begin{aligned} \sqrt{xy} &= x^2y-2\\ \frac{1}{2\sqrt{xy}}\cdot \frac{\mathrm{d}}{\mathrm{d}x}\{xy\}&=(2x)y+x^2\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}\\ \frac{y+x\frac{\mathrm{d}y}{\mathrm{d}x}}{2\sqrt{xy}}&=2xy+x^2\frac{\mathrm{d}y}{\mathrm{d}x} \\ \text{Now, we plug in $x=1$, $y=4$, and solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$:} \\ \frac{4+\frac{\mathrm{d}y}{\mathrm{d}x}}{4}&=8+\frac{\mathrm{d}y}{\mathrm{d}x}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{28}{3} \end{aligned}$$

Implicitly differentiating $x^2y(x)^2+x\sin(y(x))=4$ with respect to $x$ gives

$$\begin{aligned} 2xy^2&+2x^2yy'+\sin y+xy'\cos y=0 \\ \text{Then we gather the terms containing $y'$ on one side, so we can solve for $y'$:} \\ 2x^2yy'&+xy'\cos y=-2xy^2-\sin y\\ y'(2x^2y&+x\cos y)=-2xy^2-\sin y\\ y'&=-\frac{2xy^2+\sin y}{2x^2y+x\cos y} \end{aligned}$$