미분법 전체 문제

In the text, we saw that $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{a^x\right\}=a^x\log a$ for any constant $a$. So, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\pi^x\right\}=\pi^x\log \pi$.

By the power rule, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{x^{\pi}\right\}=\pi x^{\pi-1}$.

Therefore, $g'(x)=\pi^x\log \pi+\pi x^{\pi-1}$.

Remark: we had to use two different rules for the two different terms in $g(x)$. Although the functions $\pi^x$ and $x^\pi$ look superficially the same, they behave differently, as do their derivatives. A function of the form $(\text{constant})^{x}$ is an exponential function and not eligible for the power rule, while a function of the form $x^{\text{constant}}$ is exactly the class of function the power rule applies to.

We have the power rule to tell us the derivative of functions of the form $x^n$, where $n$ is a constant. However, here our exponent is not a constant. Similarly, in this section we learned the derivative of functions of the form $a^x$, where $a$ is a constant, but again, our base is not a constant! Although the result $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} a^x=a^x\log a$ is not what we need, the method used to differentiate $a^x$ will tell us the derivative of $x^x$.

We'll set $g(x)=\log(x^x)$, because now we can use logarithm rules to simplify:

$$\begin{aligned} g(x)=\log(f(x))&=x\log x \\ \text{Now, we can use the product rule to differentiate the right side, and the chain rule to differentiate $\log(f(x))$:} \\ g'(x)=\frac{f'(x)}{f(x)}&=\log x +x\frac{1}{x}=\log x +1 \\ \text{Finally, we solve for $f'(x)$:} \\ f'(x)&=f(x)(\log x + 1) = x^x(\log x + 1) \end{aligned}$$

We could do this with quotient and product rules, but it would be pretty painful. Insteady, let's use a logarithm.

$$\begin{aligned} f(x) &= \left(\dfrac{5x^2+10x+15}{3x^4+4x^3+5}\right)\left(\dfrac{1}{10(x+1)}\right) = \left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)\left(\dfrac{1}{2(x+1)}\right) \\ \log(f(x)) &= \log\left[\left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)\left(\dfrac{1}{2(x+1)}\right)\right]\\ &=\log\left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right) + \log\left(\dfrac{1}{2(x+1)}\right) \\&=\log\left({x^2+2x+3}\right)- \log\left({3x^4+4x^3+5}\right) -\log(x+1)-\log(2) \\ \text{Now we have a function that we can differentiate more cleanly than our original function.} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(f(x))\right\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \log\left({x^2+2x+3}\right)- \log\left({3x^4+4x^3+5}\right) - \log\left({x+1}\right)- \log\left({2}\right) \right\}\\ \frac{f'(x)}{f(x)}&=\frac{2x+2}{x^2+2x+3}-\frac{12x^3+12x^2}{3x^4+4x^3+5}-\frac{1}{x+1}\\ &=\frac{2(x+1)}{x^2+2x+3}-\frac{12x^2(x+1)}{3x^4+4x^3+5}-\frac{1}{x+1} \\ \text{Finally, we solve for $f(x)$:} \\ f'(x)&=f(x)\left(\frac{2(x+1)}{x^2+2x+3}-\frac{12x^2(x+1)}{3x^4+4x^3+5}-\frac{1}{x+1}\right)\\ &= \left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)\left(\dfrac{1}{2(x+1)}\right)\left(\frac{2(x+1)}{x^2+2x+3}-\frac{12x^2(x+1)}{3x^4+4x^3+5}-\frac{1}{x+1}\right)\\ &= \left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)\left(\frac{1}{x^2+2x+3}-\frac{6x^2}{3x^4+4x^3+5}-\frac{1}{2(x+1)^2}\right) \end{aligned}$$

We use logarithmic differentiation: we modify our function to consider

$$\begin{aligned} \log f(x) &= \log(x^2+1) \cdot \sin x \\ \text{We differentiate using the product and chain rules:} \\ \dfrac{f'(x)}{f(x)} &= \frac{\mathrm{d}}{\mathrm{d}x} \left\{ \log(x^2+1) \cdot \sin x \right\} = \cos x \cdot \log(x^2+1) + \frac{2x\sin x}{x^2+1} \\ \text{Finally, we solve for $f'(x)$} \\ f'(x) &= f(x) \cdot \left( \cos x \cdot \log(x^2+1) + \frac{2x\sin x}{x^2+1} \right) \\ &= (x^2+1)^{\sin(x)} \cdot \left( \cos x \cdot \log(x^2+1) + \frac{2x\sin x}{x^2+1} \right) \end{aligned}$$

We use logarithmic differentiation; so we modify our function to consider

$$\begin{aligned} \log f(x) &= \log(x) \cdot \cos^3(x) \\ \text{Differentiating, we find:} \\ \dfrac{f'(x)}{f(x)} &= \frac{\mathrm{d}}{\mathrm{d}x} \left\{ \log(x) \cdot \cos^3(x) \right\} = 3\cos^2(x)\cdot (-\sin(x)) \cdot \log(x) + \frac{\cos^3(x)}{x} \\ \text{Finally, we solve for $f'(x)$:} \\ f'(x) &= f(x) \cdot \left( -3\cos^2(x)\sin(x) \log(x) + \frac{\cos^3(x)}{x} \right) \\ &= x^{\cos^3(x)} \cdot \left( -3\cos^2(x)\sin(x) \log(x) + \frac{\cos^3(x)}{x} \right) \end{aligned}$$

Remark: negative numbers behave in a complicated manner when they are the base of an exponential expression. For example, the expression $(-1)^x$ is defined when $x$ is the reciprocal of an odd number (like $x=\frac{1}{5}$ or $x=\frac{1}{7}$), but not when $x$ is the reciprocal of an even number (like $x=\frac{1}{2}$). Since the domain of $f(x)$ was restricted so that $x$ is always positive, we avoid these complications.

We use logarithmic differentiation. So, we modify our function and consider

$$\begin{aligned} \log f(x) &= (x^2-3)\cdot \log(3+\sin(x))\,. \\ \text{We differentiate:} \\ \frac{f'(x)}{f(x)} &= \frac{\mathrm{d}}{\mathrm{d}x} \left\{(x^2-3)\cdot \log(3+\sin(x)) \right\}\\ &=2x\log(3+\sin(x)) + (x^2-3)\frac{\cos(x)}{3+\sin(x)} \\ \text{Finally, we solve for $f'(x)$:} \\ f'(x)&= f(x)\cdot \left[ 2x\log(3+\sin(x)) + \frac{(x^2-3)\cos(x)}{3+\sin(x)}\right]\\ &= (3+\sin(x))^{x^2-3}\cdot \left[ 2x\log(3+\sin(x)) + \frac{(x^2-3)\cos(x)}{3+\sin(x)}\right] \end{aligned}$$

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).

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 $y' = \frac{6}{25}$.