미분법 전체 문제

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

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

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

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

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:

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

We differentiate using the chain rule.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{\log(5x^2-12)\}&=\frac{10x}{5x^2-12} \\ \text{Using the quotient rule:} \\ \frac{\mathrm{d}^2}{\mathrm{d}{x}^2}\{\log(5x^2-12)\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{10x}{5x^2-12}\right\}\\ &=\frac{(5x^2-12)(10)-10x(10x)}{(5x^2-12)^2}\\ &=\frac{-10(5x^2+12)}{(5x^2-12)^2} \\ \text{Using the quotient rule one last time:} \\ \frac{\mathrm{d}^3}{\mathrm{d}{x}^3}\{\log(5x^2-12)\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \frac{-10(5x^2+12)}{(5x^2-12)^2} \right\}\\ &=\frac{(5x^2-12)^2(-10)(10x)+10(5x^2+12)(2)(5x^2-12)(10x)}{(5x^2-12)^4}\\ &=\frac{(5x^2-12)(-100x)+(200x)(5x^2+12)}{(5x^2-12)^3}\\ &=\frac{100x(-5x^2+12+10x^2+24)}{(5x^2-12)^3}\\ &=\frac{100x(5x^2+36)}{(5x^2-12)^3} \end{aligned}$$

The velocity of the particle is given by $h'(t)=\sin t$. Note $0<1<\pi$, so $h'(1)>0$–the particle is rising (moving in the positive direction, in this case “up"). The acceleration of the particle is $h''(t)=\cos t$. Since $0<1<\frac{\pi}{2}$, $h''(t)>0$, so $h'(t)$ is increasing: the particle is moving up, and it's doing so at an increasing rate. So, the particle is speeding up.

For this problem, remember that velocity has a sign indicating direction, while speed does not.

The velocity of the particle is given by $h'(t)=3t^2-2t-5$. At $t=1$, the velocity of the particle is $-4$, so the particle is moving downwards with a speed of 4 units per second. The acceleration of the particle is $h''(t)=6t-2$, so when $t=1$, the acceleration is (positive) $4$ units per second per second. That means the velocity (currently $-4$ units per second) is becoming a bigger number–since the velocity is negative, a bigger number is closer to zero, so the speed of the particle is getting smaller. (For instance, a velocity of $-3$ represents a slower motion than a velocity of $-4$.) So, the particle is slowing down at $t=1$.