미분 전체 문제

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

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.

We know that $y'=2x$. So, if we choose a point $(\alpha,\alpha^2)$ on the curve $y=x^2$, then the tangent line to the curve at that point has slope $2\alpha$. That is, the tangent line has equation

$$\begin{aligned} (y-\alpha^2)&=2\alpha(x-\alpha)\\ \text{simplified, } \qquad y&=(2\alpha)x-\alpha^2 \\ \text{So, if $(1,-3)$ is on the tangent line, then} \\ -3&=(2\alpha)(1)-\alpha^2\\ \iff\qquad 0&=\alpha^2-2\alpha-3\\ \iff\qquad 0&=(\alpha-3)(\alpha+1)\\ \iff\qquad \alpha&=3, \quad\text{or}\qquad \alpha=-1. \\ \text{So, the tangent lines $y=(2\alpha)x-\alpha^2$ are} \\ y&=6x-9 \quad\text{and}\quad y=-2x-1. \end{aligned}$$

$C'(w)=\displaystyle\lim_{h \rightarrow 0} \dfrac{C(w+h)-C(w)}{h} \approx \dfrac{C(w+1)-C(w)}{1}=C(w+1)-C(w)$, which is the number of calories in $C(w+1)$ grams minus the number of calories in $C(w)$ grams. This is the number of calories per additional gram, when there are $w$ grams.

We can use the quotient rule here.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{2x-1}{2x+1}\right\}&=\frac{(2x+1)(2)-(2x-1)(2)}{(2x+1)^2} =\frac{4}{(2x+1)^2}=\frac{1}{(x+1/2)^2} \end{aligned}$$

We already know that $\frac{\mathrm{d}}{\mathrm{d}x}x^2=2x$. So the slope of $y=3x^2$ at $x=a$ is $6a$. The tangent line to $y=3x^2$ at $x=a, y=3a^2$ is $y-3a^2=6a(x-a)$. This tangent line passes through $(2,9)$ if

$$\begin{aligned} 9-3a^2&=6a(2-a)\\ 3a^2-12a+9&=0\\ a^2-4a+3&=0\\ (a-3)(a-1)&=0\\ \implies~~a&=1,3 \end{aligned}$$

The points are $(1,3),\ (3,27)$.

This limit represents the derivative computed at $x=100180$ of the function $f(x)=\sqrt{x}$. Since the derivative of $f(x)$ is $\dfrac{1}{2\sqrt{x}}$, then its value at $x=100180$ is exactly $\dfrac{1}{2\sqrt{100180}}$.

Using the product rule, $f'(x)=(2x)g(x)+x^2g'(x)$, so $f'(0)=0\cdot g(x)+0\cdot g'(x)=0$. (Since $g$ is differentiable, $g'$ exists.)

In the quotient rule, there is a minus, not a plus. Also, $2(x+1)+2x$ is not the same as $2(x+1)$.

The correct version is:

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

$f(x)=\frac{2}{3}x^6+5x^4+12x^2+9$ is a polynomial:

$$\begin{aligned} f'(x)&=4x^5+20x^3+24x\\& =4x(x^4+5x^2+6)\\& = 4x((x^2)^2+5(x^2)+6)\\&=4x(x^2+2)(x^2+3) \end{aligned}$$

We can rewrite slightly to make every term into a power of $t$:

$$\begin{aligned} s(t)&=3t^4+5t^3-t^{-1}\\ s'(t)&=4\cdot 3t^{3}+3\cdot 5t^2-(-1)\cdot t^{-2}\\ &=12t^3+15t^2+\frac{1}{t^2} \end{aligned}$$

We could use the product rule here, but it's easier to simplify first. Don't be confused by the role reversal of $x$ and $y$: $x$ is the name of the function, and $y$ is the variable.

$$\begin{aligned} x(y) &= \left(2y+\frac{1}{y}\right)\cdot y^3\\ &=2y^4+y^2\\ x'(y)&=8y^3+2y \end{aligned}$$

We've already seen that $\frac{\mathrm{d}}{\mathrm{d}x}\{\sqrt{x}\}=\frac{1}{2\sqrt{x}}$, but if you forget this formula it is easy to figure out: $\sqrt{x}=x^{1/2}$, so $\frac{\mathrm{d}}{\mathrm{d}x}\{\sqrt{x}\}=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$.

Using the quotient rule:

$$\begin{aligned} T(x) &= \dfrac{\sqrt{x}+1}{x^2+3}\\ T'(x)&=\frac{(x^2+3)\left(\frac{1}{2\sqrt{x}}\right)-(\sqrt{x}+1)(2x)}{(x^2+3)^2} \end{aligned}$$

We use quotient rule:

$$\begin{aligned} \frac{(x^2+3)\cdot 7 - 2x\cdot (7x+2)}{(x^2+3)^2}=\frac{21 - 4x - 7x^2}{(x^2+3)^2} \end{aligned}$$

Instead of multiplying to get our usual form of this polynomial, we can use the product rule. If $f_1(x)=3x^3+4x^2+x+1$ and $f_2(x)=2x+5$, then $f_1'(x)=9x^2+8x+1$ and $f_2'(x)=2$. Then

$$\begin{aligned} f'(0)&=f_1'(0)f_2(0)+f_1(0)f_2'(0)\\ &=(1)(5)+(1)(2)=7\end{aligned}$$

Using the quotient rule, $$f'(x) = \frac{(x^2+5x)(9x^2)-(3x^3+1)(2x+5)}{(x^2+5x)^2} = \frac{3x^4+30x^3-2x-5}{(x^2+5x)^2}$$

We use quotient rule:

$$\begin{aligned} \frac{(2-x)(6x)-(3x^2+5)(-1)}{(2-x)^2}=\frac{-3x^2+12x+5}{(x-2)^2} \end{aligned}$$

We use quotient rule:

$$\begin{aligned} \frac{(3x^2+5)(-2x) - (2-x^2)(6x)}{(3x^2+5)^2}=\frac{-22x}{(3x^2+5)^2} \end{aligned}$$

We use quotient rule:

$$\begin{aligned} \frac{6x^2\cdot (x+2)-(2x^3+1)\cdot 1}{(x+2)^2}=\frac{4x^3+12x^2-1}{(x+2)^2} \end{aligned}$$

The derivative of the function is

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

The derivative is undefined if either $x<0$ or $x = 0,\pm 1$ (since the square-root is undefined for $x<0$ and the denominator is zero when $x=0,1,-1$. Putting this together — the derivative exists for $x>0, x\neq 1$.