Asosiy tarkibga o'tish

3.8-bo'lim: Giperbolik funksiyalarning hosilalari

Ushbu bobda biz ko'rib chiqmoqchi bo'lgan oxirgi funktsiyalar to'plami giperbolik funktsiyalardir. Ko'pgina jismoniy vaziyatlarda \({{\bf{e}}^x}\) va \({{\bf{e}}^{ - x}}\) kombinatsiyalari juda tez-tez paydo bo'ladi. Shu sababli bu birikmalarga nom berilgan. Oltita giperbolik funksiya mavjud va ular quyidagicha aniqlanadi.

\[\begin{array}{ll}\sinh x = \displaystyle \frac{{{{\bf{e}}^x} - {{\bf{e}}^{ - x}}}}{2} & \hspace{1.0in}\cosh x = \displaystyle \frac{{{{\bf{e}}^x} + {{\bf{e}}^{ - x}}}}{2}\\ \tanh x = \displaystyle \frac{{\sinh x}}{{\cosh x}} & \hspace{1.0in}\coth x = \displaystyle \frac{{\cosh x}}{{\sinh x}} = \displaystyle \frac{1}{{\tanh x}}\\ {\mbox{sech}}\,x = \displaystyle \frac{1}{{\cosh x}} & \hspace{1.0in}{\mbox{csch}}\,x = \displaystyle \frac{1}{{\sinh x}}\end{array}\]

Quyida uchta asosiy giperbolik funksiyaning grafiklari keltirilgan.

Graph of \(y=\cosh \left( x \right)\). It looks vaguely like an upwards opening parabola with vertex at (0,1). Graph of \(y=\sinh \left( x \right)\). It looks vaguely like an upwards like the graph of \(y=x^{3}\) starting in the third quadrant and increasing through the origin (where it flattens out briefly) then continuing to increase in the first quadrant.

Graph of \(y=\tanh \left( x \right)\). The graph starts on the left at the horizontal asymptote at \(y=-1\) and increases going through (0,0) and then approaching another horizontal asymptote at \(y=1\).

Bizda giperbolik funksiyalar haqida ham quyidagi faktlar mavjud.

\[\begin{array}{ll}\sinh \left( { - x} \right) = - \sinh \left( x \right) & \hspace{1.0in}\cosh \left( { - x} \right) = \cosh \left( x \right)\\ {\cosh ^2}\left( x \right) - {\sinh ^2}\left( x \right) = 1 & \hspace{1.0in}1 - {\tanh ^2}\left( x \right) = {\mbox{sec}}{{\mbox{h}}^{\mbox{2}}}\left( x \right)\end{array}\]

Siz shuni ta'kidlaysizki, ular ba'zi keng tarqalgan trig identifikatorlariga o'xshash, ammo unchalik bir xil emas, shuning uchun bu erda identifikatorlarni standart trig funktsiyalari bilan aralashtirib yubormaslik uchun ehtiyot bo'ling.

Giperbolik funktsiyalar eksponensial funktsiyalar nuqtai nazaridan aniqlanganligi sababli, keyingi bo'limni o'qib chiqsangiz, ularning hosilalarini topish juda oddiy. Ammo bizda yo'q, shuning uchun keyingi bo'limni ko'rib chiqqanimizdan keyin osongina isbotlanishi mumkin bo'lgan quyidagi formula kerak bo'ladi.

\[\frac{d}{{dx}}\left( {{{\bf{e}}^{ - x}}} \right) = - {{\bf{e}}^{ - x}}\]

Ushbu formula yordamida biz giperbolik sinusning hosilasini qilamiz va qolganini sizga mashq sifatida qoldiramiz.

\[\frac{d}{{dx}}\left( {\sinh x} \right) = \frac{d}{{dx}}\left( {\frac{{{{\bf{e}}^x} - {{\bf{e}}^{ - x}}}}{2}} \right) = \frac{{{{\bf{e}}^x} - \left( { - {{\bf{e}}^{ - x}}} \right)}}{2} = \frac{{{{\bf{e}}^x} + {{\bf{e}}^{ - x}}}}{2} = \cosh x\]

Qolganlari uchun biz giperbolik funktsiyaning ta'rifidan va/yoki bo'lim qoidasidan foydalanishimiz mumkin. Bu erda barcha oltita hosila mavjud.

Giperbolik funksiyalarning hosilalari

\[\begin{array}{ll}\displaystyle \frac{d}{{dx}}\left( {\sinh x} \right) = \cosh x & \hspace{1.0in} \displaystyle \frac{d}{{dx}}\left( {\cosh x} \right) = \sinh x\\ \displaystyle \frac{d}{{dx}}\left( {\tanh x} \right) = \mbox{sech}^{2}x & \hspace{1.0in} \displaystyle \frac{d}{{dx}}\left( {\coth x} \right) = - {\mbox{csch}}^{2}x\\ \displaystyle \frac{d}{{dx}}\left( {{\mbox{sech}}\,x} \right) = - {\mbox{sech }}x{\mbox{ tanh }}x & \hspace{1.0in} \displaystyle \frac{d}{{dx}}\left( {{\mbox{csch}}\,x} \right) = - {\mbox{csch }}x{\mbox{ coth }}x\end{array}\]

Bu erda giperbolik funktsiyalardan foydalangan holda bir nechta tez hosilalar mavjud.

1-misol Quyidagi funktsiyalarning har birini farqlang.
  1. \(f\left( x \right) = 2{x^5}\cosh x\)
  2. \(\displaystyle h\left( t \right) = \frac{{\sinh t}}{{t + 1}}\)
Yechim

a

\[f'\left( x \right) = 10{x^4}\cosh x + 2{x^5}\sinh x\]

b

\[h'\left( t \right) = \frac{{\left( {t + 1} \right)\cosh t - \sinh t}}{{{{\left( {t + 1} \right)}^2}}}\]