도함수의 뜻

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.

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

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.

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

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.