Раздел III. Производная и дифференциал — § 15. Производные и дифференциалы высших порядков

Кудрявцев — задача 15.13

Л. Д. Кудрявцев, «Сборник задач по математическому анализу. Том 1». Условие, подробное решение и ответ.

Задача 15.13

Найти d2yd^{2}y, считая известными du,d2u,dv,d2vdu, d^{2}u, dv, d^{2}v (дифференциал второго порядка от функции y=f(u,v)y = f(u,v)):

1) y=u(2+v)y = u(2+v);
2) y=uln⁡vy = u\ln v;
3) y=u2+v2y = \sqrt{u^{2} + v^{2}};
4) y=uv  (u>0)y = u^{v}\;(u>0).
Ответ:1) 2du dv+(2+v)d2u+ud2v;1)\ 2du\,dv + (2+v)d^{2}u + u d^{2}v;
2) 2vdu dv−uv2dv2+ln⁡v d2u+uvd2v;2)\ \dfrac{2}{v}du\,dv - \dfrac{u}{v^{2}}dv^{2} + \ln v\,d^{2}u + \dfrac{u}{v}d^{2}v;
3) v2du2−2uv du dv+u2dv2(u2+v2)3/2+ud2u+vd2vu2+v2;3)\ \dfrac{v^{2}du^{2} - 2uv\,du\,dv + u^{2}dv^{2}}{(u^{2}+v^{2})^{3/2}} + \dfrac{u d^{2}u + v d^{2}v}{\sqrt{u^{2}+v^{2}}};
4) v(v−1)uv−2du2+2uv−1(1+vln⁡u)du dv+uv(ln⁡u)2dv2+vuv−1d2u+uvln⁡u d2v4)\ v(v-1)u^{v-2}du^{2} + 2u^{v-1}(1+v\ln u)du\,dv + u^{v}(\ln u)^{2}dv^{2} + v u^{v-1}d^{2}u + u^{v}\ln u\,d^{2}v

Решение

Общая формула:d2y=∂2y∂u2du2+2∂2y∂u∂vdu dv+∂2y∂v2dv2+∂y∂ud2u+∂y∂vd2v.d^{2}y = \dfrac{\partial^{2} y}{\partial u^{2}}du^{2} + 2\dfrac{\partial^{2} y}{\partial u\partial v}du\,dv + \dfrac{\partial^{2} y}{\partial v^{2}}dv^{2} + \dfrac{\partial y}{\partial u}d^{2}u + \dfrac{\partial y}{\partial v}d^{2}v.
∂y∂u=2+v, ∂y∂v=u, ∂2y∂u2=0, ∂2y∂u∂v=1, ∂2y∂v2=0\dfrac{\partial y}{\partial u} = 2 + v,\ \dfrac{\partial y}{\partial v} = u,\ \dfrac{\partial^{2} y}{\partial u^{2}} = 0,\ \dfrac{\partial^{2} y}{\partial u\partial v} = 1,\ \dfrac{\partial^{2} y}{\partial v^{2}} = 0.d2y=2du dv+(2+v)d2u+u d2v.d^{2}y = 2du\,dv + (2+v)d^{2}u + u\,d^{2}v.
∂y∂u=ln⁡v, ∂y∂v=uv, ∂2y∂u2=0, ∂2y∂u∂v=1v, ∂2y∂v2=−uv2\dfrac{\partial y}{\partial u} = \ln v,\ \dfrac{\partial y}{\partial v} = \dfrac{u}{v},\ \dfrac{\partial^{2} y}{\partial u^{2}} = 0,\ \dfrac{\partial^{2} y}{\partial u\partial v} = \dfrac{1}{v},\ \dfrac{\partial^{2} y}{\partial v^{2}} = -\dfrac{u}{v^{2}}.d2y=2⋅1v du dv−uv2 dv2+ln⁡v d2u+uv d2v.d^{2}y = 2\cdot\dfrac{1}{v}\,du\,dv - \dfrac{u}{v^{2}}\,dv^{2} + \ln v\,d^{2}u + \dfrac{u}{v}\,d^{2}v.
∂y∂u=uu2+v2, ∂y∂v=vu2+v2, ∂2y∂u2=v2(u2+v2)3/2, ∂2y∂v2=u2(u2+v2)3/2, ∂2y∂u∂v=−uv(u2+v2)3/2\dfrac{\partial y}{\partial u} = \dfrac{u}{\sqrt{u^{2}+v^{2}}},\ \dfrac{\partial y}{\partial v} = \dfrac{v}{\sqrt{u^{2}+v^{2}}},\
\dfrac{\partial^{2} y}{\partial u^{2}} = \dfrac{v^{2}}{(u^{2}+v^{2})^{3/2}},\
\dfrac{\partial^{2} y}{\partial v^{2}} = \dfrac{u^{2}}{(u^{2}+v^{2})^{3/2}},\
\dfrac{\partial^{2} y}{\partial u\partial v} =-\dfrac{uv}{(u^{2}+v^{2})^{3/2}}.d2y=v2 du2−2uv du dv+u2 dv2(u2+v2)3/2++uu2+v2 d2u+vu2+v2 d2v.\begin{aligned}
d^{2}y &= \dfrac{v^{2}\,du^{2} - 2uv\,du\,dv + u^{2}\,dv^{2}}{(u^{2}+v^{2})^{3/2}}+ \\
&\quad + \dfrac{u}{\sqrt{u^{2}+v^{2}}}\,d^{2}u + \dfrac{v}{\sqrt{u^{2}+v^{2}}}\,d^{2}v.
\end{aligned}Логарифмируем: ln⁡y=vln⁡u\ln y = v\ln u. Тогда
∂y∂u=vuv−1, ∂y∂v=uvln⁡u.\dfrac{\partial y}{\partial u} = v u^{v-1},\ \dfrac{\partial y}{\partial v} = u^{v}\ln u.∂2y∂u2=v(v−1)uv−2,∂2y∂v2=uv(ln⁡u)2,∂2y∂u∂v=uv−1(1+vln⁡u).\dfrac{\partial^{2} y}{\partial u^{2}} = v(v-1)u^{v-2},\quad
\dfrac{\partial^{2} y}{\partial v^{2}} = u^{v}(\ln u)^{2},\quad
\dfrac{\partial^{2} y}{\partial u\partial v} = u^{v-1}(1 + v\ln u).d2y=v(v−1)uv−2 du2+2uv−1(1+vln⁡u) du dv+uv(ln⁡u)2 dv2+vuv−1 d2u+uvln⁡u d2v.\begin{aligned}
d^{2}y &= v(v-1)u^{v-2}\,du^{2} + 2u^{v-1}(1+v\ln u)\,du\,dv + u^{v}(\ln u)^{2}\,dv^{2} \\
&\quad + v u^{v-1}\,d^{2}u + u^{v}\ln u\,d^{2}v.
\end{aligned}
Кудрявцев № 15.13 — решение и ответ | Высшая математика — Профиматика