미분법 전체 문제

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.

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

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

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

Let $g(x):=\log(f(x))$. Notice $g'(x)=\frac{f'(x)}{f(x)}$.

In order to show that the two curves have horizontal tangent lines at the same values of $x$, we will show two things: first, that if $f(x)$ has a horizontal tangent line at some value of $x$, then also $g(x)$ has a horizontal tangent line at that value of $x$. Second, we will show that if $g(x)$ has a horizontal tangent line at some value of $x$, then also $f(x)$ has a horizontal tangent line at that value of $x$.

Suppose $f(x)$ has a horizontal tangent line where $x=x_0$ for some point $x_0$. This means $f'(x_0)=0$. Then $g'(x_0)=\frac{f'(x_0)}{f(x_0)}$. Since $f(x_0) \neq 0$, $\frac{f'(x_0)}{f(x_0)}=\frac{0}{f(x_0)}=0$, so $g(x)$ also has a horizontal tangent line when $x=x_0$. This shows that whenever $f$ has a horizontal tangent line, $g$ has one too.

Now suppose $g(x)$ has a horizontal tangent line where $x=x_0$ for some point $x_0$. This means $g'(x_0)=0$. Then $g'(x_0)=\frac{f'(x_0)}{f(x_0)}=0$, so $f'(x_0)$ exists and is equal to zero. Therefore, $f(x)$ also has a horizontal tangent line when $x=x_0$. This shows that whenever $g$ has a horizontal tangent line, $f$ has one too.

Remark: if we were not told that $f(x)$ gives only positive numbers, it would not necessarily be true that $f(x)$ and $\log(f(x))$ have horizontal tangent lines at the same values of $x$. If $f(x)$ had a horizontal tangent line at an $x$-value where $f(x)$ were negative, then $\log(f(x))$ would not exist there, let alone have a horizontal tangent line.