음함수의 미분법

(a) $y=-1/3x+1$ (b) $y = 3\sqrt{3}/4$

(a) $y=-\frac{1}{\sqrt{3}}(x-\frac72)+\frac{6+3\sqrt{3}}{2}$ (b) $y = \sqrt{3}(x-\frac{4+3\sqrt{3}}2)+\frac32$

(a) $y=1$ (b) $y =-\frac{2}{\sqrt{5}}(x+1)+\frac12(-1+\sqrt{5})$ (c) $y =\frac{2}{\sqrt{5}}(x+1)+\frac12(-1-\sqrt{5})$

$y'= (1+x)^{1/x}\big(\frac{1}{x(x+1)}-\frac{\ln(1+x)}{x^2}\big)$

Tangent line: $y = (1-2\ln 2)(x-1)+ 2$

$y'= (2x)^{x^2}\big(2x\ln (2x)+x\big)$

Tangent line: $y = (2+4\ln 2)(x-1)+ 2$

$y'= \frac{x^x}{x+1}\big(\ln x+1-\frac1{x+1}\big)$

Tangent line: $y = (1/4)(x-1)+ 1/2$

$y'= x^{\sin (x)+2}\big(\cos x\ln x+\frac{\sin x+2}{x}\big)$

Tangent line: $y = (3\pi^2/4)(x-\pi/2)+(\pi/2)^3$

$y'= \frac{x+1}{x+2}\big(\frac1{x+1}-\frac1{x+2}\big)$

Tangent line: $y = 1/9(x-1)+2/3$

$y'= \frac{(x+1)(x+2)}{(x+3)(x+4)}\big(\frac1{x+1}+\frac1{x+2}-\frac1{x+3}-\frac1{x+4}\big)$

Tangent line: $y = 11/72x+1/6$

We use the power rule (a) and the chain rule (b): the power rule tells us to “bring down the 2", and the chain rule tells us to multiply by $y'$.

There is no need for the quotient rule here, as there are no quotients. Exponential functions have the form $(\text{constant})^{\text{function}}$, but our function has the form $(\text{function})^{\text{constant}}$, so we did not use (d).

At $(0,4)$ and $(0,-4)$, the curve looks to be horizontal, if you zoom in: a tangent line here would have derivative zero. At the origin, the curve looks like its tangent line is vertical, so $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ does not exist.

(a) No. A function must pass the vertical line test: one input cannot result in two (or more) outputs. Since one value of $x$ sometimes corresponds to two values of $y$ (for example, when $x=\pi/4$, $y$ is $\pm 1/\sqrt{2}$), there is no function $f(x)$ so that $y=f(x)$ captures every point on the circle.

Remark: $y=\pm\sqrt{1-x^2}$ does capture every point on the unit circle. However, since one input $x$ sometimes results in two outputs $y$, this expression is not a function.

(b) No, for the same reasons as (a). If $f'(x)$ is a function, then it can give at most one slope corresponding to one value of $x$. Since one value of $x$ can correspond to two points on the circle with different slopes, $f'(x)$ cannot give the slope of every point on the circle. For example, fix any $0<a<1$. There are two points on the circle with $x$-coordinate equal to $a$. At the upper one, the slope is strictly negative. At the lower one, the slope is strictly positive.

(c) We differentiate:

$$\begin{aligned} 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{and solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{y} \end{aligned}$$

But there is a $y$ in the right-hand side of this equation, and it's not clear how to get it out. Our answer in (b) tells us that, actually, we can't get it out, if we want the right-hand side to be a function of $x$. The derivative cannot be expressed as a function of $x$, because one value of $x$ corresponds to multiple points on the circle.

Remark: since $y=\pm\sqrt{1-x^2}$, we could try writing $$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x}{y}=\pm\frac{x}{\sqrt{1-x^2}}$$ but this is not a function of $x$. Again, in a function, one input leads to at most one output, but here one value of $x$ will usually lead to two values of $\frac{\mathrm{d}y}{\mathrm{d}x}$.

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:

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