다항함수의 미분법

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 $x>0, x\neq 1$.

Using the product rule seems faster than expanding.

$$\begin{aligned} f'(x)&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{3\sqrt[5]{x}+15\sqrt[3]{x}+8\right\}\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\frac{\mathrm{d}}{\mathrm{d}x}\left\{3x^2+8x-5\right\}\\ &=\frac{\mathrm{d}}{\mathrm{d}x}\left\{3{x}^{\frac{1}{5}}+15{x}^{\frac{1}{3}}+8\right\}\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\frac{\mathrm{d}}{\mathrm{d}x}\left\{3x^2+8x-5\right\}\\ &=\left(\frac{3}{5}{x}^{\frac{-4}{5}}+5{x}^{\frac{-2}{3}}\right)\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\left(6x+8\right) \end{aligned}$$

To avoid the quotient rule, we can divide through the denominator:

$$\begin{aligned} f(x)&=\dfrac{(x^2+5x+1)(\sqrt{x}+\sqrt[3]{x})}{x} =(x^2+5x+1)\dfrac{(\sqrt{x}+\sqrt[3]{x})}{x}\\ &=(x^2+5x+1)(x^{-1/2}+x^{-2/3}) \\ \text{Now, product rule:} \\ f'(x)&=(2x+5)(x^{-1/2}+x^{-2/3})+(x^2+5x+1)\left(\frac{-1}{2}x^{-3/2}-\frac{2}{3}x^{-5/3}\right) \end{aligned}$$

(If you simplified differently, or used the quotient rule, you probably came up with a different-looking answer. There is only one derivative, though, so all correct answers will look the same after sufficient algebraic manipulation.)

Denote by $m$ the slope of the common tangent, by $(x_1,y_1)$ the point of tangency with $y=x^2$, and by $(x_2,y_2)$ the point of tangency with $y=x^2-2x+2$. Then we must have $$y_1=x_1^2\qquad y_2=x_2^2-2x_2+2\qquad m=2x_1=2x_2-2=\frac{y_2-y_1}{x_2-x_1}$$ From the “$m$” equations we get $x_1=\frac{m}{2}$, $x_2=\frac{m}{2}+1$ and

$$\begin{aligned} m&=\frac{y_2-y_1}{x_2-x_1}\\ &=y_2-y_1\\ &=x_2^2-2x_2+2-x_1^2\\ \\& =(x_2-x_1)(x_2+x_1)-2(x_2-1)\\ &=\left(\frac{m}{2}+1-\frac{m}{2}\right)\left(\frac{m}{2}+1+\frac{m}{2}\right) -2\left(\frac{m}{2}+1-1\right) \\ &=(1)(m+1)-2\frac{m}{2}\\ &=1\\ \text{So,}~~~m&=1,\qquad x_1=\tfrac{1}{2},\qquad y_1=\frac{1}{4},\qquad x_2=\frac{3}{2},\qquad y_2=\frac{9}{4}-3+2=\frac{5}{4} \end{aligned}$$

An equation of the common tangent is $y=x-\frac{1}{4}$.

The line $y=mx+b$ is tangent to $y=x^2$ at $x=\alpha$ if $$2\alpha=m\text{ and }\alpha^2=m\alpha+b \iff m=2\alpha\text{ and }b=-\alpha^2$$ The same line $y=mx+b$ is tangent to $y=-x^2+2x-5$ at $x=\beta$ if

$$\begin{aligned} -2\beta+2=m&\text{ and }-\beta^2+2\beta-5=m\beta+b\\ \iff m=2-2\beta&\text{ and }b=-\beta^2+2\beta-5-(2-2\beta)\beta=\beta^2-5 \end{aligned}$$

For the line to be simultaneously tangent to the two parabolas we need $$m=2\alpha=2-2\beta\text{ and }b=-\alpha^2=\beta^2-5$$ Substituting $\alpha=1-\beta$ into $-\alpha^2=\beta^2-5$ gives $-(1-\beta)^2=\beta^2-5$ or $2\beta^2-2\beta-4=0$ or $\beta=-1,2$. The corresponding values of the other parameters are $\alpha=2,-1$, $m=4,-2$ and $b=-4,-1$. The two lines are $y=4x-4$ and $y=-2x-1$.

This limit represents the derivative computed at $x=2$ of the function $f(x)=x^{2015}$. To see this, simply use the definition of the derivative at $a=2$ with $f(x)=x^{2015}$:

$$\begin{aligned} \left.\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}\right|_{a} &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}\\ \left.\frac{\mathrm{d}}{\mathrm{d}x}\{x^{2015}\}\right|_{2} &=\lim_{x\to2} \frac{x^{2015}-2^{2015}}{x-2} \end{aligned}$$

Since the derivative of $f(x)$ is $2015\cdot x^{2014}$, then its value at $x=2$ is exactly $2015\cdot 2^{2014}$.

(a) First, since $h(x) = 3f(x) - 4g(x)$, we know $h(2) = 3f(2) - 4g(2) = 3(5) - 4(-3) = 27$. Next, by the sum and constant multiple rules, $h'(x) = 3f'(x) - 4g'(x)$. Hence, using the given derivative values, $h'(2) = 3f'(2) - 4g'(2) = 3(-1/2) - 4(2) = -19/2$. (b) The tangent line at the point $(2,h(2))$ has equation $L(x) = h(2) + h'(2)(x-2)$. Using the information determined in (a), we see $L(x) = 27 - \frac{19}{2}(x-2)$. (c) Since $p(x) = -2f(x) + \frac{1}{2}g(x)$, it follows that $p'(x) = -2f'(x) + \frac{1}{2}g'(x)$ by the sum and constant multiple rules. Using the given function and derivative values for $f$ and $g$ at $x=2$, we see that $p(2) = -2f(2) + \frac{1}{2}g(2) = -2(5) + \frac{1}{2}(-3) = -\frac{23}{2}$ and $p'(2) = -2(-\frac{1}{2}) + \frac{1}{2}(2)) = 2$. Since $p'(2) = 2 > 0$, we conclude that $p$ is increasing at $x=2$. (d) First, the linearization of $p$ at $a = 2$ is $L(x) = p(2) + p'(2)(x-2) = -\frac{23}{2} + 2(x-2)$. Using the linearization to estimate $p(2.03)$, we find that $$p(2.03) \approx L(2.03) = -\frac{23}{2} + 2(2.03-2) = -11.5 + 2(0.03) = -11.44\,.$$