練習問題メモ 01#
次の関数を微分せよ。
\(\text { (1) } y=x^3-3 x^2+2\)
\[
\frac{dy}{dx} = 3x^2 - 6x
\]
\(\text { (2) } y=x(x-2)(x-4)(x-6)\)
\[\begin{split}
x(x-2)(x-4)(x-6)
= (x^2 - 2x)(x-4)(x-6)\\
= (x^3 - 6x^2 + 8x)(x-6)\\
= x^4 - 6x^3 - 6x^3 + 36x^2 + 8x^2 - 48x\\
= x^4 - 12x^3 + 44x^2 - 48x\\
\end{split}\]
\[
\frac{dy}{dx} = 4 x^3 - 36 x^2 + 88x - 48
\]
import sympy as sp
x = sp.symbols("x")
y = x * (x - 2) * (x - 4) * (x - 6)
sp.expand(y)
\[\displaystyle x^{4} - 12 x^{3} + 44 x^{2} - 48 x\]
sp.expand(y).diff(x)
\[\displaystyle 4 x^{3} - 36 x^{2} + 88 x - 48\]
\(\text { (3) } y=\frac{5+x}{5-x}\)
\[
y=\frac{5+x}{5-x}
= \frac{g(x)}{m(x)}
\]
とおく。商の微分公式から
\[\begin{split}
\begin{align}
\frac{dy}{dx}
&= \frac{g'(x) m(x) - g(x) m'(x)}{m(x)^2}\\
&= \frac{(5+x)' (5-x) - (5+x) (5-x)'}{(5-x)^2}\\
&= \frac{(5-x) + (5+x)}{(5-x)^2}\\
&= \frac{10}{(5-x)^2}\\
\end{align}
\end{split}\]
\(\text { (4) } y=\frac{1}{\sqrt{x^2+2}}\)
\[
y = (x^2+2)^{-1/2}
\]
\[\begin{split}
\frac{dy}{dx}
= \frac{dy}{du} \frac{du}{dx}
= - \frac{1}{2} (x^2 + 2)^{-3/2} \times 2x\\
= - \frac{ x }{x^2 + 2^{3/2}}
\end{split}\]
import sympy as sp
x = sp.symbols("x")
y = 1/sp.sqrt(x**2 + 2)
y.diff(x)
\[\displaystyle - \frac{x}{\left(x^{2} + 2\right)^{\frac{3}{2}}}\]
\(\displaystyle \text { (5) } y=\frac{\cos x}{x}\)
商の公式を使う場合
\[\begin{split}
\left(\frac{\cos x}{x}\right)'
= \frac{-\sin x \cdot x - \cos x}{x^2}\\
= \frac{-x \sin x + \cos x}{x^2}\\
\end{split}\]
積の微分公式を使う
\[\begin{split}
[\cos x \cdot x^{-1}]'
= -\sin x \cdot x^{-1} + \cos x \cdot -x^{-2}\\
= \frac{- \sin x}{x} + \frac{\cos x}{x^2}\\
= \frac{- x \sin x + \cos x}{x^2}\\
\end{split}\]
\(\text { (6) } y=e^{a x}(\cos b x+\sin b x)\)
積の公式を使って
\[\begin{split}
\begin{align}
[e^{a x}\cdot (\cos b x + \sin b x)]'
&= a e^{a x} \cdot (\cos b x + \sin b x) + e^{a x} \cdot (-b \sin bx + b \cos b x)\\
&= e^{a x} \cdot (a \cos b x + a \sin b x) + e^{a x} \cdot (-b \sin bx + b \cos b x)\\
&= e^{a x} [(a \cos b x) + (a \sin b x) -b \sin bx) + (b \cos b x)]\\
&= e^{a x} [(a +b )\cos b x + (a - b) \sin b x]\\
\end{align}
\end{split}\]
前提１：\((e^{ax})' = a e^{ax}\)
\(u=ax\)とおいて合成関数として
\[
\frac{ d e^{u} }{ d u }
\frac{ d ax }{ dx }
= e^{ax} a
\]
前提２：\((\sin bx)' = b \cos bx\)
\(u=bx\)とした合成関数として、
\[
\frac{ d \sin u }{ du }
\frac{ d u }{ dx }
= \cos bx \cdot b
\]
import sympy as sp
a, b, x = sp.symbols("a b x")
y = sp.exp(a * x) * (sp.cos(b * x) + sp.sin(b * x))
y.diff(x)
\[\displaystyle a \left(\sin{\left(b x \right)} + \cos{\left(b x \right)}\right) e^{a x} + \left(- b \sin{\left(b x \right)} + b \cos{\left(b x \right)}\right) e^{a x}\]
\(\text { (7) } y=\tan \left(x^2+2\right)\)
\(u=x^2+2\)として
\[\begin{split}
[\tan (x^2+2)]'
= \frac{dy}{du} \frac{du}{dx}\\
= \frac{1}{\cos^2 u} 2 x\\
= \frac{1}{\cos^2 (x^2+2)} 2 x\\
= \frac{ 2 x}{\cos^2 (x^2+2)}\\
\end{split}\]
\(\text { (8) } y=\log \left(\cos ^2 x\right)\)
\[\begin{split}
(\cos^2 x)'
= -\sin x \cos x + \cos x \cdot -\sin x\\
= -2 \sin x \cos x
\end{split}\]
なので
\[\begin{split}
\begin{align}
y'
&= \frac{d(\log u)}{du} \frac{du}{dx}\\
&= \frac{1}{\cos^2 x} \times -2 \sin x \cos x\\
&= -2 \frac{\sin x}{\cos x}\\
&= -2 \tan x
\end{align}
\end{split}\]
\(\text { (9) } y=\log \{\log (\log x)\}\)
\[\begin{split}
\begin{align}
\frac{dy}{dx}
&=
\frac{d (\log \{\log (\log x)\})}{d \{\log (\log x)\}}
\frac{d \{\log (\log x)\}}{d (\log x)}
\frac{d (\log x)}{dx}\\
&=
\frac{1}{\log (\log x)}
\frac{1}{\log x}
\frac{1}{x}\\
&=
\frac{1}{\log (\log x) \cdot \log x \cdot x}
\end{align}
\end{split}\]
\(\text { (10) } y=e^{x^2+3 x} \log x\)
まず\(e^{x^2+3 x}\)は合成関数の微分公式で
\[
\frac{d e^{x^2+3 x}}{d(x^2+3 x)}
\frac{d (x^2+3 x)}{dx}
= e^{x^2+3 x} ( 2x+3)
\]
\(e^{x^2+3 x} \log x\)は積の微分公式で
\[\begin{split}
\begin{align}
\frac{dy}{dx}
&= (e^{x^2+3 x})' \log x + e^{x^2+3 x} (\log x)' \\
&= e^{x^2+3 x} ( 2x+3)\log x + e^{x^2+3 x} \frac{1}{x}\\
&= e^{x^2+3 x} \left( (2x+3)\log x + \frac{1}{x} \right)\\
\end{align}
\end{split}\]
\(\text { (11) } y=8^x x^3\)
積の微分公式より
\[
(8^x)' x^3 + 8^x (x^3)'
\]
\((8^x)'\)は対数をとって逆関数の微分として解くと\((a^x)' = x \log a\)となることから
\[
(8^x)' = 8^x \log 8
\]
よって
\[\begin{split}
(8^x)' x^3 + 8^x (x^3)'
= 8^x \log 8 \cdot x^3 + 8^x 3x^2\\
= 8^x (x^3 \log 8 + 3x^2)\\
\end{split}\]
\(\displaystyle (12)\ y=\frac{x}{\sqrt{a^2-x^2}}-\arcsin \frac{x}{a}\)
まず第1項は
\[\begin{split}
\begin{align}
\left( \frac{x}{\sqrt{a^2-x^2}} \right)'
&= \frac{\sqrt{a^2-x^2} - x \frac{1}{2} (a^2-x^2)^{-1/2} \cdot (-2x) }{(\sqrt{a^2-x^2})^2}\\
&= \frac{\sqrt{a^2-x^2}}{(\sqrt{a^2-x^2})^2}
+ \frac{x^2 (a^2-x^2)^{-1/2}}{(\sqrt{a^2-x^2})^2}\\
&= \frac{1}{\sqrt{a^2-x^2}}
+ \frac{x^2}{(a^2-x^2)^{\frac{3}{2}}}\\
\end{align}
\end{split}\]
つづいて第2項は
\[
\frac{d}{d x} \arcsin x
= \frac{1}{\sqrt{1 - x^2}}
\]
より
\[\begin{split}
\begin{align}
\frac{d}{d x} \arcsin \frac{x}{a}
&= \frac{d \arcsin u}{d u} \frac{d (1/a) x}{d x}\\
&= \frac{1}{\sqrt{1 - (\frac{x}{a})^2}} \cdot \frac{1}{a}\\
&= \frac{1}{a \sqrt{1 - \frac{x^2}{a^2}}}\\
&= \frac{1}{\sqrt{a^2(1 - \frac{x^2}{a^2})}}\\
&= \frac{1}{\sqrt{a^2 - x^2}}\\
\end{align}
\end{split}\]
よって
\[
\left( \frac{x}{\sqrt{a^2-x^2}}-\arcsin \frac{x}{a}\right)'
= \frac{1}{\sqrt{a^2-x^2}}
+ \frac{x^2}{(a^2-x^2)^{\frac{3}{2}}}
- \frac{1}{\sqrt{a^2 - x^2}}
\]
整理すると
\[\begin{split}
\frac{1}{\sqrt{a^2-x^2}}
+ \frac{x^2}{(a^2-x^2)^{\frac{3}{2}}}
- \frac{1}{\sqrt{a^2 - x^2}}
\\
= \frac{ (a^2-x^2) }{ \sqrt{a^2-x^2} (a^2-x^2)}
+ \frac{x^2}{\sqrt{a^2-x^2} (a^2-x^2)}
- \frac{ (a^2-x^2)}{\sqrt{a^2 - x^2} (a^2-x^2)}
\\
= \frac{ (a^2-x^2) + x^2 - (a^2-x^2) }{ \sqrt{a^2-x^2} (a^2-x^2)}\\
= \frac{x^2}{ \sqrt{a^2-x^2} (a^2-x^2)}\\
= \frac{x^2}{(a^2-x^2)^{\frac{3}{2}}}
\end{split}\]
(13) \(\displaystyle y=\arctan \left(\frac{b}{a} \tan x\right)\)
\[\begin{split}
(\arctan(x))' = \frac{1}{x^2 + 1}\\
(\tan(x))' = \frac{1}{\cos^2 x}\\
\end{split}\]
より、
\[\begin{split}
\left[ \arctan \left(\frac{b}{a} \tan x\right)\right]'
= \frac{1}{\frac{b^2}{a^2} \tan^2 x + 1} \cdot \frac{b}{a} \frac{1}{\cos^2 x}\\
= \frac{1}{\frac{b^2}{a^2} \frac{\sin^2 x}{\cos^2 x} + 1} \cdot \frac{b}{a \cos^2 x}\\
= \frac{1}{\frac{b^2}{a^2} \frac{\sin^2 x}{\cos^2 x} + 1} \cdot \frac{ab}{a^2 \cos^2 x}\\
= \frac{ab}{(\frac{b^2}{a^2} \frac{\sin^2 x}{\cos^2 x} + 1) a^2 \cos^2 x}\\
= \frac{ab}{ b^2 \sin^2 x + a^2 \cos^2 x }\\
\end{split}\]