도함수의 뜻

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,

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

where in these inequalities we used the fact that $x^2\ge 0$. Finally, since $\lim_{x\to 0^+}-x^2=\lim_{x\to 0^+}x^2=0$, the Squeeze Theorem yields that also $\lim_{x\to 0^+}x^2\cos\left(\frac{1}{x}\right) =0$, as claimed.

Since the left and right limits match (they're both equal to $0$), we conclude that indeed $f(x)$ is differentiable at $x=0$ (and its derivative at $x=0$ is actually equal to $0$).

The function is differentiable at $x=1$ if the following limit: $$\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}\frac{f(x)-0}{x-1}=\lim_{x\to 1} \frac{f(x)}{x-1}$$ exists (note that we used the fact that $f(1)=0$ as per the definition of the first branch which includes the point $x=0$). We compute left and right limits; so $$\lim_{x\to 1^-}\frac{f(x)}{x-1}=\lim_{x\to 1^-}\frac{4x^2-8x+4}{x-1}=\lim_{x\rightarrow 1^-}\frac{4(x-1)^2}{x-1} =\lim_{x\to 1^-} 4(x-1)=0$$ and $$\lim_{x\to 1^+}\frac{(x-1)^2\sin\left(\frac{1}{x-1}\right)}{x-1}=\lim_{x\to 1^+}(x-1)\cdot \sin\left(\frac{1}{x-1}\right).$$ This last limit equals $0$ by the Squeeze Theorem since $$-1\le \sin\left(\frac{1}{x-1}\right)\le 1$$ and so, $$-(x-1)\le (x-1)\cdot \sin\left(\frac{1}{x-1}\right)\le x-1,$$ where in these inequalities we used the fact that $x\to 1^+$ yields positive values for $x-1$. Finally, since $\lim_{x\to 1^+}-x+1=\lim_{x\to 1^+}x-1=0$, the Squeeze Theorem yields that also $\lim_{x\to 1^+}(x-1)\sin\left(\frac{1}{x-1}\right) =0$, as claimed.

Since the left and right limits match (they're both equal to $0$), we conclude that indeed $f(x)$ is differentiable at $x=1$ (and its derivative at $x=1$ is actually equal to $0$).

Many answers are possible; here is one.

The key is to realize that the few points you're given suggest a pattern, but don't guarantee it. You only know nine points; anything can happen in between.