미분법 전체 문제

$\lim_{\theta \to \pi }\frac{\sin \theta }{\tan \theta }=\frac{\sin \pi }{\tan \pi }=\frac{0}{0};$ then, $\lim_{\theta \to \pi }\frac{\sin \theta }{\tan \theta }=\lim_{\theta \to \pi }\frac{\sin \theta }{\frac{\sin \theta }{\cos \theta }}=\lim_{\theta \to \pi }\cos \theta =-1$

$\frac{dy}{dx}=2x-\sec x\tan x$

$\frac{4}{{(5-2x)}^{3}}$

$2x{e}^{x}+{x}^{2}{e}^{x}$

$\frac{-3}{5}$

1

$$\frac{\mathrm{d}A}{\mathrm{d}E}=\frac{\mathrm{d}A}{\mathrm{d}B}\cdot\frac{\mathrm{d}B}{\mathrm{d}C}\cdot\frac{\mathrm{d}C}{\mathrm{d}D}\cdot\frac{\mathrm{d}D}{\mathrm{d}E}<0$$ since we multiply three positive quantities and one negative.

We are given that one speaker produces 3dB. So if $P$ is the power of one speaker,

$$\begin{aligned}3=V(P)&=10\log_{10}\left(\frac{P}{S}\right). \\ \text{So, for ten speakers:} \\ V(10P)&=10\log_{10}\left(\frac{10P}{S}\right)=10\log_{10}\left(\frac{P}{S}\right)+10\log_{10}\left(10\right)\\ &=3+10(1)=13 \mathrm{dB} \\ \text{and for one hundred speakers:} \\ V(100P)&=10\log_{10}\left(\frac{100P}{S}\right)=10\log_{10}\left(\frac{P}{S}\right)+10\log_{10}\left(100\right)\\ &=3+10(2)=23 \mathrm{dB} \end{aligned}$$

The investment doubles when it hits $2000. So, we find the value of $t$ that gives $A(t)=2000$:

$$\begin{aligned} 2000&=A(t)\\ 2000&=1000e^{t/20}\\ 2&=e^{t/20}\\ \log 2 &=\frac{t}{20}\\ 20\log2&=t \end{aligned}$$

From our logarithm rules, we know that when $y$ is positive, $\log (y^2)=2\log y$. However, the expression $\cos x$ does not always take on positive values, so (a) is not correct. (For instance, when $x=\pi$, $\log(\cos^2 x)=\log(\cos^2\pi)=\log\left((-1)^2\right) = \log(1)=0$, while $2\log (\cos \pi)=2\log(-1)$, which does not exist.)

Because $\cos^2 x$ is never negative, we notice that $\cos^2 x = |\cos x|^2$. When $\cos x$ is nonzero, $|\cos x|$ is positive, so our logarithm rules tell us $\log\left(|\cos x|^2\right) =2\log|\cos x |$. When $\cos x$ is exactly zero, then both $\log(\cos^2x)$ and $2\log|\cos x|$ do not exist. So, $\log(\cos^2x) = 2\log|\cos x|$.

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).

The derivative of $e^x$ is $e^x$: taking derivatives leaves the function unchanged, even if we do it 180 times. So $f^{(180)}=e^x$.

The derivative $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ is $\dfrac{11}{4}$ only at the point $(1,3)$: it is not constantly $\dfrac{11}{4}$, so it is wrong to differentiate the constant $\dfrac{11}{4}$ to find $\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}{x}^2}$. Below is a correct solution.

$$\begin{aligned} -28x+2y+2xy'+2yy'&=0 \\ \text{Plugging in $x=1$, $y=3$:} \\ -28+6+2y'+6y'&=0\\ y'&=\frac{11}{4} \quad\text{ at the point $(1,3)$} \\ \text{Differentiating the equation $-28x+2y+2xy'+2yy'=0$:} \\ -28+2y'+2y'+2xy''+2y'y'+2yy''&=0\\ 4y'+2(y')^2+2xy''+2yy''&=28 \\ \text{At the point $(1,3)$, $y'=\dfrac{11}{4}$. Plugging in:} \\ 4\left(\frac{11}{4}\right)+2\left(\frac{11}{4}\right)^2+2(1)y''+2(3)y''&=28\\ y''&=\frac{15}{64} \end{aligned}$$

We have an equation relating $P$ and $Q$:

$$\begin{aligned} P&=Q^3 \\ \text{We differentiate implicitly with respect to a third variable, $t$:} \\ \frac{\mathrm{d}P}{\mathrm{d}t}&=3Q^2\cdot\frac{\mathrm{d}Q}{\mathrm{d}t} \end{aligned}$$

If we know two of the three quantities $\displaystyle \frac{\mathrm{d}P}{\mathrm{d}t}$, $Q$, and $\displaystyle \frac{\mathrm{d}Q}{\mathrm{d}t}$, then we can find the third. Therefore, ii is a question we can solve. If we know $P$, then we also know $Q$ (it's just the cube root of $P$), so also we can solve iv. However, if we know neither $P$ nor $Q$, then we can't find $\displaystyle \frac{\mathrm{d}P}{\mathrm{d}t}$ based only off $\displaystyle \frac{\mathrm{d}Q}{\mathrm{d}t}$, and we can't find $\displaystyle \frac{\mathrm{d}Q}{\mathrm{d}t}$ based only off $\displaystyle \frac{\mathrm{d}P}{\mathrm{d}t}$. So we can't solve i or iii.