다항함수의 미분법

$y=-3{x}^{2}+9x-1$

We already know that $\frac{\mathrm{d}}{\mathrm{d}x}x=1$ and $\frac{\mathrm{d}}{\mathrm{d}x}x^2=2x$, so we can compute the derivative of $x^3$ by writing $x^3=(x)(x^2)$,

$$\frac{\mathrm{d}}{\mathrm{d}x} x^3 =\frac{\mathrm{d}}{\mathrm{d}x} (x)(x^2) = (1)(x^2)+(x)(2x)=3x^2$$

When this is evaluated at $x=\frac{1}{2}$ we get $\frac{3}{4}$. Since we also compute $\left( \frac{1}{2}\right)^3=\frac{1}{8}$, the equation of the tangent line is

$$\begin{aligned} y - \frac{1}{8} = \frac{3}{4}\cdot \left(x-\frac{1}{2}\right). \end{aligned}$$

We can use the quotient rule here.

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

$f(x)=\frac{2}{3}x^6+5x^4+12x^2+9$ 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 $x>0, x\neq 1$.

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.)