미분법 전체 문제

(a) $f\,'(x) = (x^2+3x) + x(2x+3)$ (b) $f\,'(x) = 3x^2 + 6x$ (c) They are equal.

(a) $g'(x) = 4x(5x^3)+2x^2(15x^2)$ (b) $g'(x) = 50x^4$ (c) They are equal.

(a) $h'(s) = 2(s+4) + (2s-1)(1)$ (b) $h'(s) = 4s + 7$ (c) They are equal.

(a) $f\,'(x) = 2x(3-x^3)+(x^2+5)(-3x^2)$ (b) $f\,'(x) = -5x^4-15x^2+6x$ (c) They are equal.

(a) $f\,'(x) = \frac{x(2x)-(x^2+3)1}{x^2}$ (b) $f\,'(x) = 1-\frac{3}{x^2}$ (c) They are equal.

(a) $g'(x) = \frac{2x^2(3x^2-4x)-(x^3-2x^2)(4x)}{4x^4}$ (b) $g'(x) = 1/2$ (c) They are equal.

(a) $h'(s) = \frac{4s^3(0) - 3(12s^2)}{16s^6}$ (b) $h'(s) = -9/4s^{-4}$ (c) They are equal.

(a) $f\,'(t) = \frac{(t+1)(2t) - (t^2-1)(1)}{(t+1)^2}$ (b) $f(t) = t-1$ when $t\neq -1$, so $f\,'(t) = 1$. (c) They are equal.

In each, $\frac{dy}{dx} = -\frac yx$.

Compose $f(g(x))$ and $g(f(x))$ to confirm that each equals $x$.

Compose $f(g(x))$ and $g(f(x))$ to confirm that each equals $x$.

Compose $f(g(x))$ and $g(f(x))$ to confirm that each equals $x$.

Compose $f(g(x))$ and $g(f(x))$ to confirm that each equals $x$.

(a) $f(x) = x$, so $f\,'(x) = 1$ (b) $f\,'(x) = \frac{1}{1+\tan^2x}\sec^2 x = 1$

First, we note that our function is continuous, because $$\lim_{x \to 0} f(x)=\lim_{x \to 0} \frac{\sin x}{x}=1=f(0)$$ This is a handy thing to check: if the function were discontinuous at $x=0$, then we would automatically know that it was not differentiable there.

Now, on to the derivative. We can use the limit definition:

$$\begin{aligned} f'(0)&=\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}\qquad\text{ if it exists}\\ &=\lim_{h \to 0} \frac{f(h)-1}{h}\\ &=\lim_{h \to 0} \frac{\frac{\sin h}{h}-1}{h} \end{aligned}$$

As $h$ approaches 0, both the numerator and the denominator approach 0. So, to evaluate the limit, we need to do more work. The key insight we can use is a result that was shown in the text while evaluating the derivative of sine. When $h$ is close to 0, $\cos h \leq \frac{\sin h}{h} \leq 1$. We use this to bound our limit, and then apply the squeeze theorem.

$$\begin{aligned} &\cos h&&\leq&&\frac{\sin h}{h}&&\leq&&1\\ \text{So, }\qquad&\cos h-1&&\leq&&\frac{\sin h}{h}-1&&\leq&& 1-1=0 \end{aligned}$$

and $\frac{1}{h}\left(\frac{\sin h}{h}-1\right)$ is between $0$ and $\frac{\cos h-1}{h}$. We can evaluate the limit of $\frac{\cos h-1}{h}$, by noticing its similarity to the definition of the derivative.

$$\begin{aligned} \lim_{h \to 0}\frac{\cos h-1}{h}&=\lim_{h \to 0}\frac{\cos(0+h)-\cos(0)}{h}\\ &=\frac{\mathrm{d}}{\mathrm{d}x}\left.\left\{\cos x \right\}\right|_{x=0}=0 \end{aligned}$$

So, by the Squeeze Theorem, $f'(0) =\displaystyle \lim_{h \to 0}\frac{\frac{\sin h}{h}-1}{h}=0$.

Let $f(x)=e^{x+x^2}$ and $g(x)=1+x$. Then $f(0)=g(0)=1$.

$f'(x)=(1+2x)e^{x+x^2}$ and $g'(x)=1$. When $x>0$, $$f'(x)=(1+2x)e^{x+x^2}>1\cdot e^{x+x^2}=e^{x+x^2}>e^{0+0^2}=1=g'(x).$$ Since $f(0)=g(0)$, and $f'(x)>g'(x)$ for all $x>0$, that means $f$ and $g$ start at the same place, but $f$ always grows faster. Therefore, $f(x)>g(x)$ for all $x>0$.

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.

(a) By definition, we know that $$f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h}\,.$$ By the sum of two angles identity for the sine function, $\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)$, and thus $$f'(x) = \lim_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$$ .. Combining the first and last terms and removing a factor of $\sin(x)$, $$f'(x) = \lim_{h \to 0} \frac{ \sin(x)(\cos(h)-1) + \cos(x)\sin(h) - \sin(x) }{h}$$ .. (b) Continuing our work in (a), we divide each part of the numerator by $h$ and see it follows $$f'(x) = \lim_{h \to 0} \left( \frac{\sin(x) (\cos(h)-1) }{h} + \frac{\cos(x)\sin(h)}{h} \right)\,.$$ Next, we note that $\sin(x)$ does not depend on $h$, and likewise $\cos(x)$ is independent of $h$. It thus makes sense to view those constants as being outside the limit being taken, and also that the limit of a sum is the sum of the limits. Hence we write $$f'(x) = \sin(x) \cdot \lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) + \cos(x) \cdot \lim_{h \to 0} \left( \frac{\sin(h)}{h} \right)\,.$$ (c) Using small positive and negative values of $h$ to estimate the desired limits, we find that it appears $$\lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) = 0$$ and $$\lim_{h \to 0} \left( \frac{\sin(h)}{h} \right) = 1$$ It suffices in each case to consider $h$-values of $h = \pm 0.01, \pm 0.001, \pm 0.0001$. (d) Combining our results in (b) and (c), we have found that $$f'(x) = \sin(x) \cdot \lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) + \cos(x) \cdot \lim_{h \to 0} \left( \frac{\sin(h)}{h} \right) = \sin(x) \cdot 0 + \cos(x) \cdot 1 = \cos(x)\,.$$ (e) The argument for showing $\frac{d}{dx}[\cos(x)] = -\sin(x)$ is almost identical. Do note that the sum of two angles identity for $\cos(\alpha + \beta)$ is $\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$.