미분법 전체 문제

• 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

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

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

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

So $x=2,-2$.

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

$$\begin{aligned} 2x+(y+1)e^y\frac{\mathrm{d}y}{\mathrm{d}x}+e^y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \end{aligned}$$

• Now set $y=0$ to get

$$\begin{aligned} 2x+(0+1)e^0\frac{\mathrm{d}y}{\mathrm{d}x}+e^0\frac{\mathrm{d}y}{\mathrm{d}x}&=0\\ 2x+\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\mathrm{d}y}{\mathrm{d}x}&=0\\ 2x&=-2\frac{\mathrm{d}y}{\mathrm{d}x}\\ x&=-\frac{\mathrm{d}y}{\mathrm{d}x} \end{aligned}$$

• So at $(x,y)=(2,0)$ we have $y' = -2$,
• and at $(x,y)=(-2,0)$ we have $y' = 2$.

The slope of the tangent line is, of course, given by the derivative, so let's start by finding $\frac{\mathrm{d}y}{\mathrm{d}x}$ of both shapes.

$$\begin{aligned} \\ \text{For the circle, we differentiate implicitly} \\ 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{and solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{y} \\ \text{For the ellipse, we also differentiate implicitly:} \\ 2x+6y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{and solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{3y} \\ \text{What we want is a value of $x$ where both derivatives are equal. However, they might have different values of $y$, so let's let $y_1$ be the $y$-values associated with $x$ on the circle, and let $y_2$ be the $y$-values associated with $x$ on the ellipse. That is, $x^2+ y_1^2=1$ and $x^2+3y_2^2=1$. For the slopes at $(x,y_1)$ on the circle and $(x,y_2)$ on the ellipse to be equal, we need:} \\ -\frac{x}{y_1}&=-\frac{x}{3y_2}\\ x\left(\frac{1}{y_1}-\frac{1}{3y_2}\right)&=0 \\ \text{So $x=0$ or $y_1=3y_2$. Let's think about which $x$-values will have a $y$-coordinate of the circle be three times as large as a $y$-coordinate of the ellipse. If $y_1=3y_2$, $(x,y_1)$ is on the circle, and $(x,y_2)$ is on the ellipse, then $x^2+y_1^2 =x^2+(3y_2)^2=1$ and $x^2+3y_2^2=1$. In this case:} \\ x^2+9y_2^2&=x^2+3y_2^2\\ 9y_2^2&=3y_2^2\\ y_2&=0\\ x&=\pm 1 \end{aligned}$$

We need to be a tiny bit careful here: when $y=0$, $y'$ is not defined for either curve. For both curves, when $y=0$, the tangent lines are vertical (and so have no real-valued slope!). Two vertical lines are indeed parallel.

So, for $x=0$ and for $x=\pm1$, the two curves have parallel tangent lines.

The derivative of $e^x$ is $e^x$: taking derivatives leaves the function unchanged, even if we do it 180 times. So $f^{(180)}=e^x$.

The derivative $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ is $\dfrac{11}{4}$ only at the point $(1,3)$: it is not constantly $\dfrac{11}{4}$, so it is wrong to differentiate the constant $\dfrac{11}{4}$ to find $\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}{x}^2}$. Below is a correct solution.

$$\begin{aligned} -28x+2y+2xy'+2yy'&=0 \\ \text{Plugging in $x=1$, $y=3$:} \\ -28+6+2y'+6y'&=0\\ y'&=\frac{11}{4} \quad\text{ at the point $(1,3)$} \\ \text{Differentiating the equation $-28x+2y+2xy'+2yy'=0$:} \\ -28+2y'+2y'+2xy''+2y'y'+2yy''&=0\\ 4y'+2(y')^2+2xy''+2yy''&=28 \\ \text{At the point $(1,3)$, $y'=\dfrac{11}{4}$. Plugging in:} \\ 4\left(\frac{11}{4}\right)+2\left(\frac{11}{4}\right)^2+2(1)y''+2(3)y''&=28\\ y''&=\frac{15}{64} \end{aligned}$$

$$\begin{aligned} f(x)&=x\log x -x\\ f'(x)&=\log x +x\cdot\frac{1}{x}-1\\ &=\log x\\ f''(x)&=\frac{1}{x} \end{aligned}$$

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

We use implicit differentiation, twice.

$$\begin{aligned} 2x+2yy'&=0\\ 2+(2y)y''+(2y')y'&=0\\ y''&=-\frac{(y')^2+1}{y} \\ \text{So, we need an expression for $y'$. We use the equation $2x+2yy'=0$ to conclude $y'=-\dfrac{x}{y}$:} \\ y''&=-\frac{\left(-\frac{x}{y}\right)^2+1}{y}\\ &=-\frac{\frac{x^2}{y^2}+1}{y}\\ &=-\frac{x^2+y^2}{y^3}\\ &=-\frac{1}{y^3} \end{aligned}$$

The question asks for $s''(1)$. We start our differentiation using the quotient rule:

$$\begin{aligned} s'(t)&=\frac{e^t(t^2+1)-e^t(2t)}{(t^2+1)^2}\\ &=\frac{e^t(t^2-2t+1)}{(t^2+1)^2} \\ \text{Using the quotient rule again,} \\ s''(t)&=\frac{(t^2+1)^2 \frac{\mathrm{d}}{\mathrm{d}t}\{e^t(t^2-2t+1)\}-e^t(t^2-2t+1) \frac{\mathrm{d}}{\mathrm{d}t}\{(t^2+1)^2\}}{(t^2+1)^4}\\ &=\frac{(t^2+1)^2\cdot\left[ e^t(2t-2)+e^t(t^2-2t+1)\right]- e^t(t^2-2t+1) \cdot 2(t^2+1)(2t)}{(t^2+1)^4}\\ &=\frac{e^t(t^2+1)^2(t^2-1)-4te^t(t-1)^2(t^2+1)}{(t^2+1)^4}\\ s''(1)&=0 \end{aligned}$$