지수함수와 로그함수의 미분

$\pi {x}^{\pi -1}\cdot {\pi }^{x}+{x}^{\pi }\cdot {\pi }^{x}\ln \pi$

${(\sin 2x)}^{4x}[4\cdot \ln (\sin 2x)+8x\cdot \cot 2x]$

Using the quotient rule,

$$\begin{aligned} f'(x)=\frac{2xe^x-2e^x}{4x^2}=\frac{e^x(2x-2)}{4x^2}=\frac{(x-1)e^x}{2x^2} \end{aligned}$$

$$\begin{aligned} f'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\{e^{2x}\}=\frac{\mathrm{d}}{\mathrm{d}x}\{(e^x)^2\}=2\frac{\mathrm{d}}{\mathrm{d}x}\{e^x\}e^x=2e^xe^x=2(e^{x})^2=2e^{2x} \end{aligned}$$

$$\begin{aligned} e^{a+x}&=e^ae^x \\ \text{Since $e^a$ is just a constant,} \\ \frac{\mathrm{d}}{\mathrm{d}x}\{e^{a}e^{x}\}&=e^a\frac{\mathrm{d}}{\mathrm{d}x}\{e^x\}=e^ae^x=e^{a+x} \end{aligned}$$

So, $f'(x)=f(x)=e^{a+x}$.

$$\begin{aligned}e^{-x}&=\frac{1}{e^x} \\ \text{Using the rule for differentiating the reciprocal:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\{e^{-x}\}&=\frac{-e^x}{(e^x)^2}=\frac{-1}{e^x}=-e^{-x} \end{aligned}$$

The question asks when $s'(t)$ is negative. So, we start by differentiating. Using the product rule:

$$\begin{aligned} s'(t)&=e^t(t^2+2t)\\ &=e^t \cdot t(t+2) \end{aligned}$$

$e^t$ is always positive, so $s'(t)$ is negative when $t$ and $2+t$ have opposite signs. This occurs when $-2<t<0$.

• Solution 1: Using the chain rule, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(10x)\right\}=\dfrac{1}{10x}\cdot 10=\frac{1}{x}$.
• Solution 2: Simplifying, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(10x)\right\}=\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(10)+\log x\right\} = 0+\frac{1}{x}=\frac{1}{x}$.

• Solution 1: Using the chain rule, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(x^2)\right\}=\frac{1}{x^2}\cdot 2x = \frac{2}{x}$.
• Solution 2: Simplifying, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(x^2)\right\}= \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{2\log(x)\right\}= \frac{2}{x}$.

Don't be fooled by a common mistake: $\log(x^2+x)$ is not the same as $\log(x^2)+\log x$.

We differentiate using the chain rule: $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(x^2+x)\right\}=\dfrac{1}{x^2+x}\cdot(2x+1) = \dfrac{2x+1}{x^2+x}$.

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!