다항함수의 미분법

Tangent line: $y = 2(x-1)$

Normal line: $y=-1/2(x-1)$

Tangent line: $y = t+4$

Normal line: $y=-t+4$

Tangent line: $y = x-1$

Normal line: $y=-x+1$

Tangent line: $y = 4$

Normal line: $x=\pi/2$

Tangent line: $y = \sqrt{2}(x-\frac{\pi}{4})-\sqrt{2}$

Normal line: $y = \frac{-1}{\sqrt{2}}(x-\frac{\pi}{4})-\sqrt{2}$

Tangent line: $y = 2x+3$

Normal line: $y=-1/2(x-5)+13$

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

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

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 $y=-2x-1$.