미분 전체 문제

$f'(-1)$ does not exist, because to the left of $x=-1$ the slope is a pretty big positive number (looks like around $+1$) and to the right the slope is $-1/4$. Since the derivative involves a limit, that limit needs to match the limit from the left and the limit from the right. The sharp angle made by the graph at $x=-1$ indicates that the left and right limits do not match, so the derivative does not exist.

$f'(3)$ also does not exist. One way to see this is to notice that the function is discontinuous here. More viscerally, note that $f(3)=1$, so as we take secant lines with one endpoint $(3,1)$, and the other endpoint just to the right of $x=3$, we get slopes that are more and more strongly negative, as shown in the picture below. If we take the limit of the slopes of these secant lines as $x$ goes to $3$ from the right, we get $-\infty$. (This certainly doesn't match the slope from the left, which is $-\frac{1}{4}$.)

At $x=-3$, there is some kind of “change" in the graph; however, it is a smooth curve, so the derivative exists here.

Using the definition of the derivative, $$s'(t)=\lim_{h \to 0}\frac{s(t+h)-s(t)}{h}$$ The units of the numerator are meters, and the units of the denominator are seconds (since the denominator comes from the change in the input of the function). So, the units of $s'(t)$ are metres per second.

Remark: we learned that the derivative of a position function gives velocity. In this example, the position is given in metres, and the velocity is measured in metres per second.

We can use point-slope form to get the equation of the line, if we have a point and its slope. The point is given: $(1,6)$. The slope is the derivative:

$$\begin{aligned} y'(1)&=\lim_{h \rightarrow 0}\frac{y(1+h)-y(1)}{h}\\ &=\lim_{h \rightarrow 0}\frac{[(1+h)^3+5]-[1^3+5]}{h}\\ &=\lim_{h \rightarrow 0}\frac{[1+3h+3h^2+h^3+5]-[1+5]}{h}\\ &=\lim_{h \rightarrow 0}\frac{3h+3h^2+h^3}{h}\\ &=\lim_{h \rightarrow 0}{3+3h+h^2}\\ &=3 \end{aligned}$$

So our slope is 3, which gives a line of equation $y-6=3(x-1)$.

We set up the definition of the derivative.

$$\begin{aligned} f'(x)&=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h \rightarrow 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}\\ &=\lim_{h \rightarrow 0}\frac{\frac{x}{x(x+h)}-\frac{x+h}{x(x+h)}}{h}\\ &=\lim_{h \rightarrow 0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}\\ &=\lim_{h \rightarrow 0}\frac{\frac{-h}{x(x+h)}}{h}\\ &=\lim_{h \rightarrow 0}\frac{-1}{x(x+h)}\\ &=\frac{-1}{x^2} \end{aligned}$$

By definition $$f'(0)=\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} =\lim_{h\rightarrow 0}\frac{h|h|}{h} =\lim_{h\rightarrow 0}|h|=0$$ In particular, the limit exists, so the derivative exists (and is equal to zero).

We set up the definition of the derivative.

$$\begin{aligned} f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} =\lim_{h\rightarrow 0}\frac{1}{h}\Big(\frac{2}{x+h+1} -\frac{2}{x+1}\Big) =\lim_{h\rightarrow 0}\frac{2}{h}\ \frac{(x+1)-(x+h+1)}{(x+h+1)(x+1)}\\ &=\lim_{h\rightarrow 0}\frac{2}{h}\ \frac{-h}{(x+h+1)(x+1)} =\lim_{h\rightarrow 0}\frac{-2}{(x+h+1)(x+1)} =\frac{-2}{(x+1)^2} \end{aligned}$$

$$\begin{aligned} f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} =\lim_{h\rightarrow 0}\frac{1}{h}\Big(\frac{1}{(x+h)^2+3} -\frac{1}{x^2+3}\Big) =\lim_{h\rightarrow 0}\frac{1}{h}\frac{x^2-(x+h)^2}{[(x+h)^2+3][x^2+3]}\\ &=\lim_{h\rightarrow 0}\frac{1}{h}\frac{-2xh-h^2}{[(x+h)^2+3][x^2+3]} =\lim_{h\rightarrow 0}\frac{-2x-h}{[(x+h)^2+3][x^2+3]} =\boxed{\frac{-2x}{[x^2+3]^2}} \end{aligned}$$

The slope of the tangent line is the derivative. We set this up using the same definition of the derivative that we always do. This limit is hard to take for general $x$, but easy when $x=0$.

$$\begin{aligned} f'(0)&=\lim_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\\ &=\lim_{h \rightarrow 0} \frac{h\log_{10}(2h+10)-0}{h}\\ &=\lim_{h \rightarrow 0} \log_{10}(2h+10)=\log_{10}(10)=1 \end{aligned}$$

So, the slope of the tangent line is 1.

$$\begin{aligned} f'(x)&=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} =\lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h} =\lim_{h\rightarrow 0}\frac{x^2-(x+h)^2}{(x+h)^2x^2h} =\lim_{h\rightarrow 0}\frac{-2xh-h^2}{(x+h)^2x^2h}\\ &=\lim_{h\rightarrow 0}\frac{-2x-h}{(x+h)^2x^2}=\frac{-2x}{x^4} =-\frac{2}{x^3} \end{aligned}$$

When $x$ is not equal to 2, then the function is differentiable– the only place we have to worry about is when $x$ is exactly 2.

In order for $f$ to be differentiable at $x=2$, it must also be continuous at $x=2$. This forces $x^2\big|_{x=2}=\big[ax+b\big]_{x=2}$ or $$2a+b=4.$$

In order for a limit to exist, the left- and right-hand limits must exist and be equal to each other. Since a derivative is a limit, in order for $f$ to be differentiable at $x=2$, the left hand derivative of $ax+b$ at $x=2$ must be the same as the right hand derivative of $x^2$ at $x=2$. Since $ax+b$ is a line, its derivative is $a$ everywhere. We've already seen the derivative of $x^2$ is $2x$, so we need $$a=2x\big|_{x=2}=4.$$

So, the values of $a$ and $b$ that makes $f$ differentiable everywhere are $a=4$ and $b=-4$.

We plug in $f(x)$ to the definition of a derivative. To evaluate the limit, we multiply and divide by the conjugate of the numerator, then simplify.

$$\begin{aligned} f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} &=\lim\limits_{h\rightarrow 0}\frac{\sqrt{1+x+h}-\sqrt{1+x}}{h}\\ &=\lim\limits_{h\rightarrow 0}\frac{\sqrt{1+x+h}-\sqrt{1+x}}{h} \left(\frac{\sqrt{1+x+h}+\sqrt{1+x}}{\sqrt{1+x+h}+\sqrt{1+x}}\right)\\ &=\lim\limits_{h\rightarrow 0}\frac{(1+x+h)-(1+x)}{h(\sqrt{1+x+h}+\sqrt{1+x})}\\ &=\lim\limits_{h\rightarrow 0}\frac{h}{h(\sqrt{1+x+h}+\sqrt{1+x})}\\ &=\lim\limits_{h\rightarrow 0}\frac{1}{\sqrt{1+x+h}+\sqrt{1+x}}\\ &=\frac{1}{\sqrt{1+x+0}+\sqrt{1+x}}=\frac{1}{2\sqrt{1+x}} \end{aligned}$$

The domain of the function is $[-1,\infty)$. In particular, $f(x)$ is defined when $x=-1$. However, $f'(x)$ is not defined when $x=-1$, so $f'(x)$ only exists over $(-1,\infty)$.

Remark: $\displaystyle \lim_{x \to -1^+}f'(x)=\infty$, so the tangent line to $f(x)$ at the point $x=-1$ has a vertical slope.

The function is differentiable at $x=0$ if the following limit: $$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)-0}{x}=\lim_{x\to 0} \frac{f(x)}{x}$$ exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$). We start by computing the left limit. For this computation, recall that if $x<0$ then $\sqrt{x^2}=|x|=-x$. $$\lim_{x\to 0^-}\frac{f(x)}{x}=\lim_{x\to 0^-}\frac{\sqrt{x^2+x^4}}{x}=\lim_{x\to 0^-} \frac{\sqrt{x^2}\sqrt{1+x^2}}{x}=\lim_{x \rightarrow 0}\frac{-x\sqrt{1+x^2}}{x}=-1$$ Now, from the right: $$\lim_{x\to 0^+}\frac{x\cos x}{x}=\lim_{x\to 0^+}\cos x = 1.$$ Since the limit from the left does not equal the limit from the right, the derivative does not exist at $x=0$.

The function is differentiable at $x=0$ if the following limit: $$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)-0}{x}=\lim_{x\to 0} \frac{f(x)}{x}$$ exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$).

We start by computing the left limit.

$$\begin{aligned} \lim_{x\to 0^-}\frac{f(x)}{x}=\lim_{x\to 0^-} \frac{x\cos x}{x} =\lim_{x\to 0^-} \cos x = 1. \end{aligned}$$

Now, from the right:

$$\begin{aligned} \lim_{x\to 0^+}\frac{\sqrt{1+x}-1}{x} &= \lim_{x\to 0^+}\frac{\sqrt{1+x}-1}{x} \cdot \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} \\ &= \lim_{x\to 0^+} \frac{1+x-1}{x(\sqrt{1+x}+1)} = \lim_{x\to 0^+} \frac{1}{\sqrt{1+x}+1} = \frac{1}{2} \end{aligned}$$

Since the limit from the left does not equal the limit from the right, the derivative does not exist at $x=0$.

The function is differentiable at $x=0$ if the following limit:

$$\begin{aligned} \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)-0}{x}=\lim_{x\to 0} \frac{f(x)}{x} \end{aligned}$$

exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$). We compute left and right limits; so

$$\begin{aligned} \lim_{x\to 0^-}\frac{f(x)}{x}=\lim_{x\to 0^-}\frac{x^3-7x^2}{x}=\lim_{x\to 0^-} x^2-7x=0 \end{aligned}$$

and

$$\begin{aligned} \lim_{x\to 0^+}\frac{x^3\cos\left(\frac{1}{x}\right)}{x}=\lim_{x\to 0^+}x^2\cdot \cos\left(\frac{1}{x}\right). \end{aligned}$$

This last limit equals $0$ by the Squeeze Theorem since

$$\begin{aligned} -1\le \cos\left(\frac{1}{x}\right)\le 1 \end{aligned}$$

and so,

$$\begin{aligned} -x^2\le x^2\cdot \cos\left(\frac{1}{x}\right)\le x^2, \end{aligned}$$