미분법 전체 문제

We know the derivative of the natural logarithm (base $e$), so we use the base-change formula:

$$\begin{aligned} f(x)=\log_{10}x&=\frac{\log x}{\log 10} \\ \text{Since $\log 10$ is a constant:} \\ f'(x)&=\frac{1}{x\log 10}. \end{aligned}$$

• Solution 1: Using the quotient rule, $$y'=\frac{x^3\frac{1}{x}-(\log x)\cdot 3x^2}{x^6}= \frac{x^2-3x^2\log x}{x^6}=\frac{1-3\log x}{x^4}.$$
• Solution 2: Using the product rule with $y=\log x \cdot x^{-3}$, $$y'=\frac{1}{x}x^{-3}+\log x \cdot(-3)x^{-4}=x^{-4}(1-3\log x)$$

Using the chain rule,

$$\begin{aligned} \displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta} \log(\sec \theta)&=\frac{1}{\sec \theta}\cdot (\sec \theta \cdot \tan \theta)\\ &=\tan\theta \end{aligned}$$

Remark: the domain of the function $\log(\sec \theta)$ is those values of $\theta$ for which $\sec\theta$ is positive: so, the intervals $\left(\left(2n-\frac{1}{2}\right)\pi,\left(2n+\frac{1}{2}\right)\pi\right)$ where $n$ is any integer. Certainly the tangent function has a larger domain than this, but outside the domain of $\log(\sec \theta)$, $\tan \theta$ is not the derivative of $\log(\sec \theta)$.

Let's start in with the chain rule.

$$\begin{aligned} f'(x) &= e^{ \cos\left(\log x\right)} \cdot \frac{\mathrm{d}}{\mathrm{d}x} \left\{ \cos\left( \log x \right)\right\} \\ \text{We'll need the chain rule again:} \\ &= e^{\cos\left(\log x\right)} (-\sin( \log x)) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ \log x \}\\ &= e^{\cos\left(\log x\right)} (-\sin(\log x)) \cdot \frac{1}{x}\\ &=\frac{-e^{\cos(\log x)}\sin(\log x)}{x} \end{aligned}$$

Remark: Although we have a logarithm in the exponent, we can't cancel. The expression $e^{\cos (\log x)}$ is not the same as the expression $x^{\cos x}$, or $\cos x$.

$$\begin{aligned} y&=\log(x^2+\sqrt{x^4+1}) \\ \text{So, we'll need the chain rule:} \\ y'&=\frac{\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^2+\sqrt{x^4+1}\right\}}{ x^2+\sqrt{x^4+1}}\\ &=\frac{2x+\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^4+1}\right\}}{x^2+\sqrt{x^4+1}} \\ \text{We need the chain rule again:} \\ &=\frac{2x+\frac{\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^4+1\right\}}{2\sqrt{ x^4+1}}}{x^2+\sqrt{x^4+1}}\\ &=\frac{2x+\frac{4x^3}{2\sqrt{x^4+1}}}{x^2+\sqrt{x^4+1}}. \end{aligned}$$

This requires us to apply the chain rule twice.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \left\{ \sqrt{-\log(\cos x)} \right\} &= \frac{1}{2\sqrt{ -\log(\cos x)}} \cdot \frac{\mathrm{d}}{\mathrm{d}x} \left\{ -\log\left(\cos x\right) \right\}\\ &= -\frac{1}{2\sqrt{-\log(\cos x)}} \cdot \frac{1}{ \cos x} \frac{\mathrm{d}}{\mathrm{d}x} \left\{ \cos x \right\}\\ &= -\frac{1}{2\sqrt{-\log(\cos x)}} \cdot \frac{1}{\cos x} \cdot \left(-\sin x\right)\\ &=\frac{\tan x}{2\sqrt{-\log(\cos x)}} \end{aligned}$$

Remark: it looks strange to see a negative sign in the argument of a square root. Since the cosine function always gives values that are at most 1, $\log(\cos x)$ is always negative or zero over its domain. So, $\sqrt{\log(\cos x)}$ is only defined for the points where $\cos x=1$ (and so $\log(\cos x) = 0$–this isn't a very interesting function! In contrast, $-\log(\cos x)$ is always positive or zero over its domain – and therefore we can always take its square root.

Under the chain rule, $\frac{\mathrm{d}}{\mathrm{d}x}\log f(x)=\frac{1}{f(x)}f'(x)$. So

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log\big(x+\sqrt{x^2+4}\big)\right\} &=\frac{1}{ x+\sqrt{x^2+4}} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\left\{ x+\sqrt{x^2+4}\right\} \\ &=\frac{1}{x+\sqrt{x^2+4}}\cdot\left(1+\frac{2x}{2\sqrt{x^2+4}}\right)\\ &=\frac{1}{ x+\sqrt{x^2+4}}\cdot \left(\frac{ 2 \sqrt{x^2+4} + 2 x} { 2\sqrt{x^2+4}}\right)\\ &=\frac{1}{\sqrt{x^2+4}} \end{aligned}$$

Using the chain rule,

$$\begin{aligned} g'(x)&=\frac{\frac{\mathrm{d}}{\mathrm{d}x}\{e^{x^2}+\sqrt{1+x^4}\}}{e^{x^2}+\sqrt{1+x^4}}\\ &=\frac{2xe^{x^2}+\frac{4x^3}{2\sqrt{1+x^4}}}{e^{x^2}+\sqrt{1+x^4}}\left(\frac{\sqrt{1+x^4}}{\sqrt{1+x^4}}\right)\\ &=\frac{2xe^{x^2}\sqrt{1+x^4}+2x^3}{e^{x^2}\sqrt{1+x^4}+1+x^4} \end{aligned}$$

Using logarithm rules makes this an easier problem:

$$\begin{aligned} g(x) &= \log(2x-1) -\log(2x+1)\\ \text{So, }~ g'(x) &= \dfrac{2}{2x-1} -\dfrac{2}{2x+1}\\ \text{and }~ g'(1) &= \dfrac{2}{1} -\dfrac{2}{3}=\dfrac{4}{3} \end{aligned}$$

We begin by simplifying:

$$\begin{aligned} f(x) &= \log\left(\sqrt{\dfrac{(x^2+5)^3}{x^4+10}}\right)\\ &=\log\left(\left({\dfrac{(x^2+5)^3}{x^4+10}}\right)^{1/2}\right)\\ &=\frac{1}{2}\log\left(\dfrac{(x^2+5)^3}{x^4+10}\right)\\ &=\frac{1}{2}\left[\log\left({(x^2+5)^3}\right)-\log({x^4+10})\right] \\ &=\frac{1}{2}\left[3\log\left({(x^2+5)}\right)-\log({x^4+10})\right] \\ \text{Now, we differentiate using the chain rule:} \\ f'(x)&=\frac{1}{2}\left[ 3\frac{2x}{x^2+5}-\frac{4x^3}{x^4+10} \right]\\ &=\frac{3x}{x^2+5}-\frac{2x^3}{x^4+10} \end{aligned}$$

Remark: it is a common mistake to write $\log(x^2+4)$ as $\log(x^2)+\log(4)$. These expressions are not equivalent!

We use the chain rule twice, followed by the product rule:

$$\begin{aligned} f'(x) &= \frac{1}{ g(xh(x))}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ g(xh(x))\}\\ &=\frac{1}{g(xh(x))}\cdot g'( xh(x))\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ xh(x)\} \\ &=\dfrac{1}{g\big(xh(x)\big)}\cdot g'\big(xh(x)\big)\cdot\big[h(x)+xh'(x)\big] \\ \text{In particular, when $x=2$:} \\ f'(2) &= \dfrac{1}{g\big(2h(2)\big)}\cdot g'\big(2h(2)\big)\cdot \big[h(2)+2h'(2)\big]\\ &= \dfrac{g'(4)}{g(4)}\big[2+2\times 3\big] = \dfrac{5}{3}\big[2+2\times 3\big]\\ &=\dfrac{40}{3} \end{aligned}$$

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.

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