【1987-45-4 分】 已知 z = arctan x + y x − y \displaystyle z=\arctan \dfrac{x+y}{x-y} z = arctan x − y x + y d z \displaystyle d z d z
【1988-5-4 分】 已知 u = e x y \displaystyle u=e^{\frac{x}{y}} u = e y x
【1989-5-5 分】 已知 z = a x 2 − y 2 \displaystyle z=a^{\sqrt{x^{2}-y^{2}}} z = a x 2 − y 2 a > 0 \displaystyle a>0 a > 0 a ≠ 1 \displaystyle a \ne1 a = 1 d z \displaystyle d z d z
【1991-45-3 分】 设 z = e sin x y \displaystyle z=e^{\sin x y} z = e s i n x y d z = \displaystyle d z= d z =
【1994-45-5 分】 已知 f ( x , y ) = x 2 arctan y x − y 2 arctan x y \displaystyle f(x, y)=x^{2} \arctan \dfrac{y}{x}-y^{2} \arctan \dfrac{x}{y} f ( x , y ) = x 2 arctan x y − y 2 arctan y x ∂ 2 f ∂ x ∂ y \displaystyle \dfrac{\partial^{2} f}{\partial x \partial y} ∂ x ∂ y ∂ 2 f
【1994-12-3 分】 设 u = e − x sin x y \displaystyle u=e^{-x} \sin \dfrac{x}{y} u = e − x sin y x ∂ 2 u ∂ x ∂ y \displaystyle \dfrac{\partial^{2} u}{\partial x \partial y} ∂ x ∂ y ∂ 2 u ( 2 , 1 π ) \displaystyle (2, \dfrac{1}{\pi}) ( 2 , π 1 )
【1998-34-5 分】 设 z = ( x 2 + y 2 ) e − arctan y x \displaystyle z=(x^{2}+y^{2}) e^{-\arctan \frac{y}{x}} z = ( x 2 + y 2 ) e − a r c t a n x y d z \displaystyle d z d z ∂ 2 z ∂ x ∂ y \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y} ∂ x ∂ y ∂ 2 z
【2005-34-4 分】 设二元函数 z = x e x + y + ( x + 1 ) ln ( 1 + y ) \displaystyle z=x e^{x+y}+(x+1) \ln (1+y) z = x e x + y + ( x + 1 ) ln ( 1 + y ) d z ∣ ( 1 , 0 ) = \displaystyle d z|_{(1,0)}= d z ∣ ( 1 , 0 ) =
【2008-2-4 分】 设 z = ( y x ) x y \displaystyle z=(\dfrac{y}{x})^{\frac{x}{y}} z = ( x y ) y x ∂ z ∂ x ∣ ( 1 , 2 ) = \displaystyle \dfrac{\partial z}{\partial x}|_{(1,2)}= ∂ x ∂ z ∣ ( 1 , 2 ) =
【2008-23-4 分】 设函数 f 连续, 若 F ( u , v ) = ∬ D u v f ( x 2 + y 2 ) x 2 + y 2 d x d y \displaystyle F(u, v)=\iint_{D_{u v}} \dfrac{f(x^{2}+y^{2})}{\sqrt{x^{2}+y^{2}}} ~d x ~d y F ( u , v ) = ∬ D uv x 2 + y 2 f ( x 2 + y 2 ) d x d y D u v \displaystyle D_{u v} D uv ∂ F ∂ u = ( ) \displaystyle \dfrac{\partial F}{\partial u}=( ) ∂ u ∂ F = ( ) v f ( u 2 ) \displaystyle v f(u^{2}) v f ( u 2 ) v u f ( u 2 ) \displaystyle \dfrac{v}{u} f(u^{2}) u v f ( u 2 ) v f ( u ) \displaystyle v f(u) v f ( u ) v u f ( u ) \displaystyle \dfrac{v}{u} f(u) u v f ( u )
【2009-3-4 分】 设 z = ( x + e y ) x \displaystyle z=(x+e^{y})^{x} z = ( x + e y ) x ∂ z ∂ x ∣ ( 1 , 0 ) = \displaystyle \dfrac{\partial z}{\partial x}|_{(1,0)}= ∂ x ∂ z ∣ ( 1 , 0 ) =
【2011-1-4 分】 设函数 F ( x , y ) = ∫ 0 x y sin t 1 + t 2 d t \displaystyle F(x, y)=\int_{0}^{x y} \dfrac{\sin t}{1+t^{2}} d t F ( x , y ) = ∫ 0 x y 1 + t 2 sin t d t ∂ 2 F ∂ x 2 ∣ x = 0 y = 2 = \displaystyle \dfrac{\partial^{2} F}{\partial x^{2}}|_{\substack{x=0 \\ y=2}}= ∂ x 2 ∂ 2 F ∣ x = 0 y = 2 =
【2011-3-4 分】 设函数 z = ( 1 + x y ) x y \displaystyle z=(1+\dfrac{x}{y})^{\frac{x}{y}} z = ( 1 + y x ) y x d z ∣ ( 1 , 1 ) = \displaystyle d z|_{(1,1)}= d z ∣ ( 1 , 1 ) =
【2016-23-4 分】 已知函数 f ( x , y ) = e x x − y \displaystyle f(x, y)=\dfrac{e^{x}}{x-y} f ( x , y ) = x − y e x f x ′ − f y ′ = 0 \displaystyle f_{x}'-f_{y}'=0 f x ′ − f y ′ = 0 f x ′ + f y ′ = 0 \displaystyle f_{x}'+f_{y}'=0 f x ′ + f y ′ = 0 f x ′ − f y ′ = f \displaystyle f_{x}'-f_{y}'=f f x ′ − f y ′ = f f x ′ + f y ′ = f \displaystyle f_{x}'+f_{y}'=f f x ′ + f y ′ = f
【2017-23-4 分】 设函数 f ( x , y ) \displaystyle f(x, y) f ( x , y ) d f ( x , y ) = y e y d x + x ( 1 + y ) e y d y \displaystyle d f(x, y)=y e^{y} ~d x +x(1+y) e^{y} ~d y df ( x , y ) = y e y d x + x ( 1 + y ) e y d y f ( 0 , 0 ) = 0 \displaystyle f(0,0)=0 f ( 0 , 0 ) = 0 f ( x , y ) = \displaystyle f(x, y)= f ( x , y ) =
【2020-1-4 分】 设函数 f ( x , y ) = ∫ 0 x y e x t 2 d t \displaystyle f(x, y)=\int_{0}^{x y} e^{x t^{2}} d t f ( x , y ) = ∫ 0 x y e x t 2 d t ∂ 2 f ∂ x ∂ y ∣ ( 1 , 1 ) = \displaystyle \dfrac{\partial^{2} f}{\partial x \partial y}|_{(1,1)}= ∂ x ∂ y ∂ 2 f ∣ ( 1 , 1 ) =
【2020-23-4 分】 设 z = arctan [ x y + sin ( x + y ) ] \displaystyle z=\arctan [x y+\sin (x+y)] z = arctan [ x y + sin ( x + y )] d z ∣ ( 0 , π ) = \displaystyle d z|_{(0, \pi)}= d z ∣ ( 0 , π ) =
【2021-23-5 分】 设函数 f ( x , y ) \displaystyle f(x, y) f ( x , y ) f ( x + 1 , e x ) = x ( x + 1 ) 2 \displaystyle f(x+1, e^{x})=x(x+1)^{2} f ( x + 1 , e x ) = x ( x + 1 ) 2 f ( x , x 2 ) = 2 x 2 ⋅ ln x \displaystyle f(x, x^{2})=2 x^{2} \cdot \ln x f ( x , x 2 ) = 2 x 2 ⋅ ln x d f ( 1 , 1 ) = ( ) \displaystyle d f(1,1)=( ) df ( 1 , 1 ) = ( ) d x + d y \displaystyle d x+d y d x + d y d x − d y \displaystyle d x-d y d x − d y d y \displaystyle d y d y − d y \displaystyle -d y − d y
【2022-23-5 分】 已知 f ( x ) \displaystyle f(x) f ( x ) F ( x , y ) = ∫ 0 x − y ( x − y − t ) f ( t ) d t \displaystyle F(x, y)=\int_{0}^{x-y}(x-y-t) f(t) d t F ( x , y ) = ∫ 0 x − y ( x − y − t ) f ( t ) d t ∂ F ∂ x = ∂ F ∂ y , ∂ 2 F ∂ x 2 = ∂ 2 F ∂ y 2 \displaystyle \dfrac{\partial F}{\partial x}=\dfrac{\partial F}{\partial y}, \dfrac{\partial^{2} F}{\partial x^{2}}=\dfrac{\partial^{2} F}{\partial y^{2}} ∂ x ∂ F = ∂ y ∂ F , ∂ x 2 ∂ 2 F = ∂ y 2 ∂ 2 F ∂ F ∂ x = ∂ F ∂ y , ∂ 2 F ∂ x 2 = − ∂ 2 F ∂ y 2 \displaystyle \dfrac{\partial F}{\partial x}=\dfrac{\partial F}{\partial y}, \dfrac{\partial^{2} F}{\partial x^{2}}=-\dfrac{\partial^{2} F}{\partial y^{2}} ∂ x ∂ F = ∂ y ∂ F , ∂ x 2 ∂ 2 F = − ∂ y 2 ∂ 2 F ∂ F ∂ x = − ∂ F ∂ y , ∂ 2 F ∂ x 2 = ∂ 2 F ∂ y 2 \displaystyle \dfrac{\partial F}{\partial x}=-\dfrac{\partial F}{\partial y}, \dfrac{\partial^{2} F}{\partial x^{2}}=\dfrac{\partial^{2} F}{\partial y^{2}} ∂ x ∂ F = − ∂ y ∂ F , ∂ x 2 ∂ 2 F = ∂ y 2 ∂ 2 F ∂ F ∂ x = − ∂ F ∂ y , ∂ 2 F ∂ x 2 = − ∂ 2 F ∂ y 2 \displaystyle \dfrac{\partial F}{\partial x}=-\dfrac{\partial F}{\partial y}, \dfrac{\partial^{2} F}{\partial x^{2}}=-\dfrac{\partial^{2} F}{\partial y^{2}} ∂ x ∂ F = − ∂ y ∂ F , ∂ x 2 ∂ 2 F = − ∂ y 2 ∂ 2 F
【2023-3-5 分】 已知函数 f ( x , y ) \displaystyle f(x, y) f ( x , y ) d f ( x , y ) = x d y − y d x x 2 + y 2 \displaystyle d f(x, y)=\dfrac{x ~d y-y ~d x}{x^{2}+y^{2}} df ( x , y ) = x 2 + y 2 x d y − y d x f ( 1 , 1 ) = π 4 \displaystyle f(1,1)=\dfrac{\pi}{4} f ( 1 , 1 ) = 4 π f ( 3 , 3 ) = \displaystyle f(\sqrt{3}, 3)= f ( 3 , 3 ) =
【2024-1-5 分】 已知 f ( u , v ) \displaystyle f(u, v) f ( u , v ) d f ( 1 , 1 ) = 3 d u + 4 d v \displaystyle d f(1,1)=3 ~d u+4 d v df ( 1 , 1 ) = 3 d u + 4 d v y = f ( cos x , 1 + x 2 ) \displaystyle y = f(\cos x,1+x^{2}) y = f ( cos x , 1 + x 2 ) d y d x ∣ x = 0 = \displaystyle \dfrac{d y}{d x}|_{x=0} = d x d y ∣ x = 0 =
【1987-1-3 分】 设函数 f , g 连续可微, u = f ( x , x y ) \displaystyle u=f(x, x y) u = f ( x , x y ) v = g ( x + x y ) \displaystyle v=g(x+x y) v = g ( x + x y ) ∂ u ∂ x \displaystyle \dfrac{\partial u}{\partial x} ∂ x ∂ u ∂ v ∂ x \displaystyle \dfrac{\partial v}{\partial x} ∂ x ∂ v
【1989-12-5 分】 设 z = f ( 2 x − y ) + g ( x , x y ) \displaystyle z=f(2 x-y)+g(x, x y) z = f ( 2 x − y ) + g ( x , x y ) f ( t ) \displaystyle f(t) f ( t ) g ( u , v ) \displaystyle g(u, v) g ( u , v ) ∂ 2 z ∂ x ∂ y \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y} ∂ x ∂ y ∂ 2 z
【1992-45-5 分】 设 z = sin ( x y ) + φ ( x , x y ) \displaystyle z=\sin (x y)+\varphi(x, \dfrac{x}{y}) z = sin ( x y ) + φ ( x , y x ) ∂ 2 z ∂ x ∂ y \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y} ∂ x ∂ y ∂ 2 z φ ( u , v ) \displaystyle \varphi(u, v) φ ( u , v )
【1992-12-5 分】 设 z = f ( e x sin y , x 2 + y 2 ) \displaystyle z=f(e^{x} \sin y, x^{2}+y^{2}) z = f ( e x sin y , x 2 + y 2 ) ∂ 2 z ∂ x ∂ y \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y} ∂ x ∂ y ∂ 2 z
【1995-45-3 分】 设 z = x y f ( y x ) , f ( u ) \displaystyle z=x y f(\dfrac{y}{x}), f(u) z = x y f ( x y ) , f ( u ) x z x ′ + y z y ′ = \displaystyle x z_{x}'+y z_{y}'= x z x ′ + y z y ′ =
【1995-12-5 分】 设 u = f ( x , y , z ) \displaystyle u=f(x, y, z) u = f ( x , y , z ) φ ( x 2 , e y , z ) = 0 \displaystyle \varphi(x^{2}, e^{y}, z)=0 φ ( x 2 , e y , z ) = 0 y = sin x \displaystyle y=\sin x y = sin x ∂ φ ∂ z ≠ 0 \displaystyle \dfrac{\partial \varphi}{\partial z} \ne0 ∂ z ∂ φ = 0 d u d x \displaystyle \dfrac{d u}{~d x} d x d u
【1996-4-5 分】 设函数 z = f ( u ) \displaystyle z = f(u) z = f ( u ) u = φ ( u ) + ∫ y x p ( t ) d t \displaystyle u = \varphi(u) + \int_{y}^{x} p(t) d t u = φ ( u ) + ∫ y x p ( t ) d t f ( u ) \displaystyle f(u) f ( u ) φ ( u ) \displaystyle \varphi(u) φ ( u ) p ( t ) \displaystyle p(t) p ( t ) φ ′ ( u ) \displaystyle \varphi'(u) φ ′ ( u ) φ ′ ( u ) ≠ 1 \displaystyle \varphi'(u) \ne1 φ ′ ( u ) = 1 p ( y ) ∂ z ∂ x + p ( x ) ∂ z ∂ y \displaystyle p(y) \dfrac{\partial z}{\partial x}+p(x) \dfrac{\partial z}{\partial y} p ( y ) ∂ x ∂ z + p ( x ) ∂ y ∂ z
【1997-34-5 分】 设 u = f ( x , y , z ) \displaystyle u=f(x, y, z) u = f ( x , y , z ) y = y ( x ) \displaystyle y=y(x) y = y ( x ) z = z ( x ) \displaystyle z=z(x) z = z ( x ) e x y − y = 0 \displaystyle e^{x y}-y=0 e x y − y = 0 e z − x z = 0 \displaystyle e^{z}-x z=0 e z − x z = 0 d u d x \displaystyle \dfrac{d u}{~d x} d x d u
【1998-1-3 分】 设 z = 1 x f ( x y ) + y φ ( x + y ) \displaystyle z=\dfrac{1}{x} f(x y)+y \varphi(x+y) z = x 1 f ( x y ) + y φ ( x + y ) ∂ 2 z ∂ x ∂ y = \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y}= ∂ x ∂ y ∂ 2 z =
【2000-3-3 分】 设 z = f ( x y , x y ) + g ( y x ) \displaystyle z=f(x y, \dfrac{x}{y})+g(\dfrac{y}{x}) z = f ( x y , y x ) + g ( x y ) ∂ z ∂ x = \displaystyle \dfrac{\partial z}{\partial x}= ∂ x ∂ z =
【2000-1-5 分】 设 z = f ( x y , x y ) + g ( y x ) \displaystyle z=f(x y, \dfrac{x}{y})+g(\dfrac{y}{x}) z = f ( x y , y x ) + g ( x y ) ∂ 2 z ∂ x ∂ y \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y} ∂ x ∂ y ∂ 2 z
【2001-34-5 分】 设 u = f ( x , y , z ) \displaystyle u=f(x, y, z) u = f ( x , y , z ) z = z ( x ) \displaystyle z=z(x) z = z ( x ) e x y − x y = 2 \displaystyle e^{x y}-x y=2 e x y − x y = 2 e x = ∫ 0 x − z sin t t d t \displaystyle e^{x}=\int_{0}^{x-z} \dfrac{\sin t}{t} d t e x = ∫ 0 x − z t sin t d t d u d x \displaystyle \dfrac{d u}{~d x} d x d u
【2001-4-3 分】 设 z = e − x − f ( x − 2 y ) \displaystyle z=e^{-x}-f(x-2 y) z = e − x − f ( x − 2 y ) y = 0 \displaystyle y=0 y = 0 z = x 2 \displaystyle z=x^{2} z = x 2 ∂ z ∂ x = \displaystyle \dfrac{\partial z}{\partial x}= ∂ x ∂ z =
【2004-3-4 分】 函数 f ( u , v ) \displaystyle f(u, v) f ( u , v ) f [ x g ( y ) , y ] = x + g ( y ) \displaystyle f[x g(y), y]=x+g(y) f [ xg ( y ) , y ] = x + g ( y ) g ( y ) \displaystyle g(y) g ( y ) g ( y ) ≠ 0 \displaystyle g(y) \ne0 g ( y ) = 0 ∂ 2 f ∂ u ∂ v = \displaystyle \dfrac{\partial^{2} f}{\partial u \partial v}= ∂ u ∂ v ∂ 2 f =
【2005-12-4 分】 设函数 u ( x , y ) = φ ( x + y ) + φ ( x − y ) + ∫ x − y x + y ψ ( t ) d t \displaystyle u(x, y)=\varphi(x+y)+\varphi(x-y)+\int_{x-y}^{x+y} \psi(t) d t u ( x , y ) = φ ( x + y ) + φ ( x − y ) + ∫ x − y x + y ψ ( t ) d t ∂ 2 u ∂ x 2 = − ∂ 2 u ∂ y 2 \displaystyle \dfrac{\partial^{2} u}{\partial x^{2}}=-\dfrac{\partial^{2} u}{\partial y^{2}} ∂ x 2 ∂ 2 u = − ∂ y 2 ∂ 2 u ∂ 2 u ∂ x 2 = ∂ 2 u ∂ y 2 \displaystyle \dfrac{\partial^{2} u}{\partial x^{2}}=\dfrac{\partial^{2} u}{\partial y^{2}} ∂ x 2 ∂ 2 u = ∂ y 2 ∂ 2 u ∂ 2 u ∂ x ∂ y = ∂ 2 u ∂ y 2 \displaystyle \dfrac{\partial^{2} u}{\partial x \partial y}=\dfrac{\partial^{2} u}{\partial y^{2}} ∂ x ∂ y ∂ 2 u = ∂ y 2 ∂ 2 u ∂ 2 u ∂ x ∂ y = ∂ 2 u ∂ x 2 \displaystyle \dfrac{\partial^{2} u}{\partial x \partial y}=\dfrac{\partial^{2} u}{\partial x^{2}} ∂ x ∂ y ∂ 2 u = ∂ x 2 ∂ 2 u
【2006-3-4 分】 设函数 f ( u ) \displaystyle f(u) f ( u ) f ′ ( 0 ) = 1 2 \displaystyle f'(0)=\dfrac{1}{2} f ′ ( 0 ) = 2 1 z = f ( 4 x 2 − y 2 ) \displaystyle z=f(4 x^{2}-y^{2}) z = f ( 4 x 2 − y 2 ) d z ∣ ( 1 , 2 ) = \displaystyle d z|_{(1,2)}= d z ∣ ( 1 , 2 ) =
【2007-234-4 分】 设 f ( u , v ) \displaystyle f(u, v) f ( u , v ) z = f ( y x , x y ) \displaystyle z=f(\dfrac{y}{x}, \dfrac{x}{y}) z = f ( x y , y x ) x ∂ z ∂ x − y ∂ z ∂ y = \displaystyle x \dfrac{\partial z}{\partial x}-y \dfrac{\partial z}{\partial y}= x ∂ x ∂ z − y ∂ y ∂ z =
【2007-1-4 分】 设 f ( u , v ) \displaystyle f(u, v) f ( u , v ) z = f ( x y , y x ) \displaystyle z=f(x^{y}, y^{x}) z = f ( x y , y x ) ∂ z ∂ x = \displaystyle \dfrac{\partial z}{\partial x}= ∂ x ∂ z =
【2009-1-4 分】 设函数 f ( u , v ) \displaystyle f(u, v) f ( u , v ) z = f ( x , x y ) \displaystyle z=f(x, x y) z = f ( x , x y ) ∂ 2 z ∂ x ∂ y = \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y}= ∂ x ∂ y ∂ 2 z =
【2012-2-4 分】 设 z = f ( ln x + 1 y ) \displaystyle z=f(\ln x+\dfrac{1}{y}) z = f ( ln x + y 1 ) f ( u ) \displaystyle f(u) f ( u ) x ∂ z ∂ x + y 2 ∂ z ∂ y = \displaystyle x \dfrac{\partial z}{\partial x}+y^{2} \dfrac{\partial z}{\partial y}= x ∂ x ∂ z + y 2 ∂ y ∂ z =
【2013-2-4 分】 设 z = y x f ( x y ) \displaystyle z=\dfrac{y}{x} f(x y) z = x y f ( x y ) x y ∂ z ∂ x + ∂ z ∂ y = ( ) \displaystyle \dfrac{x}{y} \dfrac{\partial z}{\partial x}+\dfrac{\partial z}{\partial y}=( ) y x ∂ x ∂ z + ∂ y ∂ z = ( ) 2 y f ′ ( x y ) \displaystyle 2 y f'(x y) 2 y f ′ ( x y ) − 2 y f ′ ( x y ) \displaystyle -2 y f'(x y) − 2 y f ′ ( x y ) 2 x f ( x y ) \displaystyle \dfrac{2}{x} f(x y) x 2 f ( x y ) − 2 x f ( x y ) \displaystyle -\dfrac{2}{x} f(x y) − x 2 f ( x y )
【2015-2-4 分】 设函数 f ( u , v ) \displaystyle f(u, v) f ( u , v ) f ( x + y , y x ) = x 2 − y 2 \displaystyle f(x+y, \dfrac{y}{x})=x^{2}-y^{2} f ( x + y , x y ) = x 2 − y 2 ∂ f ∂ u ∣ u = 1 , v = 1 \displaystyle \dfrac{\partial f}{\partial u}|_{u=1, v=1} ∂ u ∂ f ∣ u = 1 , v = 1 ∂ f ∂ v ∣ u = 1 , v = 1 \displaystyle \dfrac{\partial f}{\partial v}|_{u=1, v=1} ∂ v ∂ f ∣ u = 1 , v = 1 1 2 , 0 \displaystyle \dfrac{1}{2}, 0 2 1 , 0 0 , 1 2 \displaystyle 0, \dfrac{1}{2} 0 , 2 1 − 1 2 , 0 \displaystyle -\dfrac{1}{2}, 0 − 2 1 , 0 0 , − 1 2 \displaystyle 0,-\dfrac{1}{2} 0 , − 2 1
【2016-13-4 分】 设函数 f ( u , v ) \displaystyle f(u, v) f ( u , v ) z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) ( x + 1 ) z − y 2 = x 2 f ( x − z , y ) \displaystyle (x+1) z-y^{2}=x^{2} f(x-z, y) ( x + 1 ) z − y 2 = x 2 f ( x − z , y ) d z ∣ ( 0 , 1 ) = \displaystyle d z|_{(0,1)}= d z ∣ ( 0 , 1 ) =
【2019-2-4 分】 设函数 f ( u ) \displaystyle f(u) f ( u ) z = y f ( y 2 x ) \displaystyle z=y f(\dfrac{y^{2}}{x}) z = y f ( x y 2 ) 2 x ∂ z ∂ x + y ∂ z ∂ y = \displaystyle 2 x \dfrac{\partial z}{\partial x}+y \dfrac{\partial z}{\partial y}= 2 x ∂ x ∂ z + y ∂ y ∂ z =
【2019-1-4 分】 设函数 f ( u ) \displaystyle f(u) f ( u ) z = f ( sin y − sin x ) + x y \displaystyle z=f(\sin y-\sin x)+x y z = f ( sin y − sin x ) + x y 1 cos x ∂ z ∂ x + 1 cos y ∂ z ∂ y = \displaystyle \dfrac{1}{\cos x} \dfrac{\partial z}{\partial x}+\dfrac{1}{\cos y} \dfrac{\partial z}{\partial y}= cos x 1 ∂ x ∂ z + cos y 1 ∂ y ∂ z =
【2022-1-5 分】 设 f ( u ) \displaystyle f(u) f ( u ) z = x y f ( y x ) \displaystyle z=x y f(\dfrac{y}{x}) z = x y f ( x y ) x ∂ z ∂ x + y ∂ z ∂ y = x y ( ln y − ln x ) \displaystyle x \dfrac{\partial z}{\partial x}+y \dfrac{\partial z}{\partial y}=x y(\ln y-\ln x) x ∂ x ∂ z + y ∂ y ∂ z = x y ( ln y − ln x ) f ( 1 ) = 1 2 , f ′ ( 1 ) = 0 \displaystyle f(1)=\dfrac{1}{2}, f'(1)=0 f ( 1 ) = 2 1 , f ′ ( 1 ) = 0 f ( 1 ) = 0 , f ′ ( 1 ) = 1 2 \displaystyle f(1)=0, f'(1)=\dfrac{1}{2} f ( 1 ) = 0 , f ′ ( 1 ) = 2 1 f ( 1 ) = 1 , f ′ ( 1 ) = 0 \displaystyle f(1)=1, f'(1)=0 f ( 1 ) = 1 , f ′ ( 1 ) = 0 f ( 1 ) = 0 , f ′ ( 1 ) = 1 \displaystyle f(1)=0, f'(1)=1 f ( 1 ) = 0 , f ′ ( 1 ) = 1
【1988-4-4 分】 已知 u + e u = x y \displaystyle u+e^{u}=x y u + e u = x y ∂ 2 u ∂ x ∂ y \displaystyle \dfrac{\partial^{2} u}{\partial x \partial y} ∂ x ∂ y ∂ 2 u
【1990-5-5 分】 设 x 2 + z 2 = y φ ( z y ) \displaystyle x^{2}+z^{2}=y \varphi(\dfrac{z}{y}) x 2 + z 2 = y φ ( y z ) ∂ z ∂ y \displaystyle \dfrac{\partial z}{\partial y} ∂ y ∂ z
【1991-5-5 分】 已知 x y = x f ( z ) + y g ( z ) \displaystyle x y=x f(z)+y g(z) x y = x f ( z ) + y g ( z ) x f ′ ( z ) + y g ′ ( z ) ≠ 0 \displaystyle x f'(z)+y g'(z) \ne0 x f ′ ( z ) + y g ′ ( z ) = 0 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) [ x − g ( z ) ] ∂ z ∂ x = [ y − f ( z ) ] ∂ z ∂ y \displaystyle [x-g(z)] \dfrac{\partial z}{\partial x}=[y-f(z)] \dfrac{\partial z}{\partial y} [ x − g ( z )] ∂ x ∂ z = [ y − f ( z )] ∂ y ∂ z
【1991-12-3 分】 由方程 x y z + x 2 + y 2 + z 2 = 2 \displaystyle x y z+\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{2} x y z + x 2 + y 2 + z 2 = 2 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) d z = \displaystyle d z= d z =
【1993-45-5 分】 设 z = f ( x , y ) \displaystyle z=f(x, y) z = f ( x , y ) z − y − x + x e z − y − x = 0 \displaystyle z-y-x+x e^{z-y-x}=0 z − y − x + x e z − y − x = 0
【1999-1-5 分】 设 y = y ( x ) \displaystyle y = y(x) y = y ( x ) z = z ( x ) \displaystyle z = z(x) z = z ( x ) z = x f ( x + y ) \displaystyle z = x f(x+y) z = x f ( x + y ) F ( x , y , z ) = 0 \displaystyle F(x,y,z) = 0 F ( x , y , z ) = 0 d z d x \displaystyle \dfrac{d z}{d x} d x d z
【1999-4-3 分】 设 f ( x , y , z ) = e x y z 2 \displaystyle f(x, y, z)=e^{x} y z^{2} f ( x , y , z ) = e x y z 2 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) x + y + z + x y z = 0 \displaystyle x+y+z+x y z=0 x + y + z + x y z = 0 f x ′ ( 0 , 1 , − 1 ) = \displaystyle f_{x}'(0,1,-1)= f x ′ ( 0 , 1 , − 1 ) =
【2004-2-4 分】 设函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) z = e 2 x − 3 z + 2 y \displaystyle z=e^{2 x-3 z}+2 y z = e 2 x − 3 z + 2 y 3 ∂ z ∂ x + ∂ z ∂ y = \displaystyle 3 \dfrac{\partial z}{\partial x}+\dfrac{\partial z}{\partial y}= 3 ∂ x ∂ z + ∂ y ∂ z =
【2005-1-4 分】 设有三元方程 x y − z ln y + e x z = 1 \displaystyle x y-z \ln y+e^{x z}=1 x y − z ln y + e x z = 1 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) y = y ( x , z ) \displaystyle y=y(x, z) y = y ( x , z ) z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) x = x ( y , z ) \displaystyle x=x(y, z) x = x ( y , z ) z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) x = x ( y , z ) \displaystyle x=x(y, z) x = x ( y , z ) y = y ( x , z ) \displaystyle y=y(x, z) y = y ( x , z )
【2010-12-4 分】 设函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) F ( y x , z x ) = 0 \displaystyle F(\dfrac{y}{x}, \dfrac{z}{x})=0 F ( x y , x z ) = 0 F 2 ′ ≠ 0 \displaystyle F_{2}' \ne0 F 2 ′ = 0 x ∂ z ∂ x + y ∂ z ∂ y = ( ) \displaystyle x \dfrac{\partial z}{\partial x}+y \dfrac{\partial z}{\partial y}=( ) x ∂ x ∂ z + y ∂ y ∂ z = ( ) x \displaystyle x x z \displaystyle z z − x \displaystyle -x − x − z \displaystyle -z − z
【2013-3-4 分】 设函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) ( z + y ) x = x y \displaystyle (z+y)^{x}=x y ( z + y ) x = x y ∂ z ∂ x ∣ ( 1 , 2 ) = \displaystyle \dfrac{\partial z}{\partial x}|_{(1,2)}= ∂ x ∂ z ∣ ( 1 , 2 ) =
【2014-2-4 分】 设 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) e 2 y z + x + y 2 + z = 7 4 \displaystyle e^{2 y z}+x+y^{2}+z=\dfrac{7}{4} e 2 y z + x + y 2 + z = 4 7 d z ∣ ( 1 2 , 1 2 ) = \displaystyle d z|_{(\frac{1}{2}, \frac{1}{2})}= d z ∣ ( 2 1 , 2 1 ) =
【2015-23-4 分】 若函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) e x + 2 y + 3 z + x y z = 1 \displaystyle e^{x+2 y+3 z}+x y z=1 e x + 2 y + 3 z + x y z = 1 d z ∣ ( 0 , 0 ) = \displaystyle d z|_{(0,0)}= d z ∣ ( 0 , 0 ) =
【2015-1-4 分】 若函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) e z + x y z + x + cos x = 2 \displaystyle e^{z}+x y z+x+\cos x=2 e z + x y z + x + cos x = 2 d z ∣ ( 0 , 1 ) = \displaystyle d z|_{(0,1)}= d z ∣ ( 0 , 1 ) =
【2018-2-4 分】 设函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) ln z + e z − 1 = x y \displaystyle \ln z+e^{z-1}=x y ln z + e z − 1 = x y ∂ z ∂ x ∣ ( 2 , 1 2 ) = \displaystyle \dfrac{\partial z}{\partial x}|_{(2, \frac{1}{2})}= ∂ x ∂ z ∣ ( 2 , 2 1 ) =
【2021-2-5 分】 设函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) ( x + 1 ) ⋅ z + y ln z − arctan ( 2 x y ) = 1 \displaystyle (x+1) \cdot z+y \ln z-\arctan (2 x y)=1 ( x + 1 ) ⋅ z + y ln z − arctan ( 2 x y ) = 1 ∂ z ∂ x ∣ ( 0 , 2 ) = \displaystyle \dfrac{\partial z}{\partial x}|_{(0,2)}= ∂ x ∂ z ∣ ( 0 , 2 ) =
【2023-2-5 分】 设函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) e z + x z = 2 x − y \displaystyle e^{z}+x z=2 x-y e z + x z = 2 x − y ∂ 2 z ∂ x 2 ∣ ( 1 , 1 ) = \displaystyle \dfrac{\partial^{2} z}{\partial x^{2}}|_{(1,1)}= ∂ x 2 ∂ 2 z ∣ ( 1 , 1 ) =
【2024-3-5 分】 设函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) z + e x − y ln ( 1 + z 2 ) = 0 \displaystyle z+e^{x}-y \ln (1+z^{2})=0 z + e x − y ln ( 1 + z 2 ) = 0 ( ∂ 2 z ∂ x 2 + ∂ 2 z ∂ y 2 ) ∣ ( 0 , 0 ) \displaystyle \left.\left(\dfrac{\partial^{2} z}{\partial x^{2}}+\dfrac{\partial^{2} z}{\partial y^{2}}\right)\right|_{(0,0)} ( ∂ x 2 ∂ 2 z + ∂ y 2 ∂ 2 z ) ( 0 , 0 )
【2025-2-5 分】 设函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) z + ln z − ∫ y x e − t 2 d t = 0 \displaystyle z+\ln z-\int_{y}^{x} e^{-t^{2}} d t=0 z + ln z − ∫ y x e − t 2 d t = 0 ∂ z ∂ x + ∂ z ∂ y = （） \displaystyle \dfrac{\partial z}{\partial x}+\dfrac{\partial z}{\partial y}=（ ） ∂ x ∂ z + ∂ y ∂ z = （） z z + 1 ( e − x 2 − e − y 2 ) \displaystyle \dfrac{z}{z+1}(e^{-x^{2}}-e^{-y^{2}}) z + 1 z ( e − x 2 − e − y 2 ) z z + 1 ( e − x 2 + e − y 2 ) \displaystyle \dfrac{z}{z+1}(e^{-x^{2}}+e^{-y^{2}}) z + 1 z ( e − x 2 + e − y 2 ) − z z + 1 ( e − x 2 − e − y 2 ) \displaystyle -\dfrac{z}{z+1}(e^{-x^{2}}-e^{-y^{2}}) − z + 1 z ( e − x 2 − e − y 2 ) − z z + 1 ( e − x 2 + e − y 2 ) \displaystyle -\dfrac{z}{z+1}(e^{-x^{2}}+e^{-y^{2}}) − z + 1 z ( e − x 2 + e − y 2 )
【2025-3-5 分】 已知函数 z = z ( x , y ) \displaystyle z=z(x, y) z = z ( x , y ) z + ln z − ∫ y x x e − t 2 d t = 1 \displaystyle z+\ln z-\int_{y}^{x} x e^{-t^{2}} d t=1 z + ln z − ∫ y x x e − t 2 d t = 1 ∂ 2 z ∂ x 2 ∣ ( 1 , 1 ) = \displaystyle \dfrac{\partial^{2} z}{\partial x^{2}}|_{(1,1)}= ∂ x 2 ∂ 2 z ∣ ( 1 , 1 ) =
【1987-2-7 分】 设函数 z = f ( u , x , y ) \displaystyle z=f(u, x, y) z = f ( u , x , y ) u = x e y \displaystyle u=x e^{y} u = x e y ∂ 2 z ∂ x ∂ y \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y} ∂ x ∂ y ∂ 2 z
【1988-12-6 分】 设 u = y f ( x y ) + x g ( y x ) \displaystyle u=y f(\dfrac{x}{y})+x g(\dfrac{y}{x}) u = y f ( y x ) + xg ( x y ) x ∂ 2 u ∂ x 2 + y ∂ 2 u ∂ x ∂ y \displaystyle x \dfrac{\partial^{2} u}{\partial x^{2}}+y \dfrac{\partial^{2} u}{\partial x \partial y} x ∂ x 2 ∂ 2 u + y ∂ x ∂ y ∂ 2 u
【1993-2-6 分】 设 z = x 3 f ( x y , y x ) \displaystyle z=x^{3} f(x y, \dfrac{y}{x}) z = x 3 f ( x y , x y ) ∂ z ∂ y \displaystyle \dfrac{\partial z}{\partial y} ∂ y ∂ z ∂ 2 z ∂ y 2 \displaystyle \dfrac{\partial^{2} z}{\partial y^{2}} ∂ y 2 ∂ 2 z ∂ 2 z ∂ x ∂ y \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y} ∂ x ∂ y ∂ 2 z
【1996-12-6 分】 设变换 { u = x − 2 y v = x + a y \displaystyle \begin{cases}u=x-2 y \\ v=x+a y\end{cases} { u = x − 2 y v = x + a y 6 ∂ 2 z ∂ x 2 + ∂ 2 z ∂ x ∂ y − ∂ 2 z ∂ y 2 = 0 \displaystyle 6 \dfrac{\partial^{2} z}{\partial x^{2}}+\dfrac{\partial^{2} z}{\partial x \partial y}-\dfrac{\partial^{2} z}{\partial y^{2}}=0 6 ∂ x 2 ∂ 2 z + ∂ x ∂ y ∂ 2 z − ∂ y 2 ∂ 2 z = 0 ∂ 2 z ∂ u ∂ v = 0 \displaystyle \dfrac{\partial^{2} z}{\partial u \partial v}=0 ∂ u ∂ v ∂ 2 z = 0
【2001-1-6 分】 设函数 z = f ( x , y ) \displaystyle z=f(x, y) z = f ( x , y ) f ( 1 , 1 ) = 1 \displaystyle f(1,1)=1 f ( 1 , 1 ) = 1 ∂ f ∂ x ∣ ( 1 , 1 ) = 2 \displaystyle \dfrac{\partial f}{\partial x}|_{(1,1)}=2 ∂ x ∂ f ∣ ( 1 , 1 ) = 2 ∂ f ∂ y ∣ ( 1 , 1 ) = 3 \displaystyle \dfrac{\partial f}{\partial y}|_{(1,1)}=3 ∂ y ∂ f ∣ ( 1 , 1 ) = 3 φ ( x ) = f ( x , f ( x , x ) ) \displaystyle \varphi(x)=f(x, f(x, x)) φ ( x ) = f ( x , f ( x , x )) d d x φ 3 ( x ) ∣ x = 1 \displaystyle \dfrac{d}{~d x} \varphi^{3}(x)|_{x=1} d x d φ 3 ( x ) ∣ x = 1
【2003-34-8 分】 设 f ( u , v ) \displaystyle f(u, v) f ( u , v ) ∂ 2 f ∂ u 2 + ∂ 2 f ∂ v 2 = 1 \displaystyle \dfrac{\partial^{2} f}{\partial u^{2}}+\dfrac{\partial^{2} f}{\partial v^{2}}=1 ∂ u 2 ∂ 2 f + ∂ v 2 ∂ 2 f = 1 g ( x , y ) = f [ x y , 1 2 ( x 2 − y 2 ) ] \displaystyle g(x, y)=f[x y, \dfrac{1}{2}(x^{2}-y^{2})] g ( x , y ) = f [ x y , 2 1 ( x 2 − y 2 )] ∂ 2 g ∂ x 2 + ∂ 2 g ∂ y 2 \displaystyle \dfrac{\partial^{2} g}{\partial x^{2}}+\dfrac{\partial^{2} g}{\partial y^{2}} ∂ x 2 ∂ 2 g + ∂ y 2 ∂ 2 g
【2004-2-10 分】 设 z = f ( x 2 − y 2 , e x y ) \displaystyle z=f(x^{2}-y^{2}, e^{x y}) z = f ( x 2 − y 2 , e x y ) ∂ z ∂ x \displaystyle \dfrac{\partial z}{\partial x} ∂ x ∂ z ∂ z ∂ y \displaystyle \dfrac{\partial z}{\partial y} ∂ y ∂ z ∂ 2 z ∂ x ∂ y \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y} ∂ x ∂ y ∂ 2 z
【2005-34-8 分】 设 f ( u ) \displaystyle f(u) f ( u ) g ( x , y ) = f ( y x ) + y f ( x y ) \displaystyle g(x, y)=f(\dfrac{y}{x})+y f(\dfrac{x}{y}) g ( x , y ) = f ( x y ) + y f ( y x ) x 2 ∂ 2 g ∂ x 2 − y 2 ∂ 2 g ∂ y 2 \displaystyle x^{2} \dfrac{\partial^{2} g}{\partial x^{2}}-y^{2} \dfrac{\partial^{2} g}{\partial y^{2}} x 2 ∂ x 2 ∂ 2 g − y 2 ∂ y 2 ∂ 2 g
【2009-2-10 分】 设 z = f ( x + y , x − y , x y ) \displaystyle z=f(x+y, x-y, x y) z = f ( x + y , x − y , x y ) d z \displaystyle d z d z ∂ 2 z ∂ x ∂ y \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y} ∂ x ∂ y ∂ 2 z
【2010-2-11 分】 设函数 u = f ( x , y ) \displaystyle u=f(x, y) u = f ( x , y ) 4 ∂ 2 u ∂ x 2 + 12 ∂ 2 u ∂ x ∂ y + 5 ∂ 2 u ∂ y 2 = 0 \displaystyle 4 \dfrac{\partial^{2} u}{\partial x^{2}}+12 \dfrac{\partial^{2} u}{\partial x \partial y}+5 \dfrac{\partial^{2} u}{\partial y^{2}}=0 4 ∂ x 2 ∂ 2 u + 12 ∂ x ∂ y ∂ 2 u + 5 ∂ y 2 ∂ 2 u = 0 ξ = x + a y \displaystyle \xi=x+a y ξ = x + a y η = x + b y \displaystyle \eta=x+b y η = x + b y ∂ 2 u ∂ ξ ∂ η = 0 \displaystyle \dfrac{\partial^{2} u}{\partial \xi \partial \eta}=0 ∂ ξ ∂ η ∂ 2 u = 0
【2011-3-10 分】 已知函数 f ( u , v ) \displaystyle f(u, v) f ( u , v ) f ( 1 , 1 ) = 2 \displaystyle f(1,1)=2 f ( 1 , 1 ) = 2 f ( u , v ) \displaystyle f(u,v) f ( u , v ) z = f ( x + y , f ( x , y ) ) \displaystyle z = f(x+y,f(x,y)) z = f ( x + y , f ( x , y )) ∂ z ∂ x ∂ y ∣ ( 1 , 1 ) \displaystyle \dfrac{\partial z}{\partial x \partial y}|_{(1,1)} ∂ x ∂ y ∂ z ∣ ( 1 , 1 )
【2011-12-10 分】 设函数 z = f ( x y , y g ( x ) ) \displaystyle z=f(x y, y g(x)) z = f ( x y , y g ( x )) g ( x ) \displaystyle g(x) g ( x ) x = 1 \displaystyle x=1 x = 1 g ( 1 ) = 1 \displaystyle g(1)=1 g ( 1 ) = 1 ∂ 2 z ∂ x ∂ y ∣ x = 1 \displaystyle \dfrac{\partial^{2} z}{\partial x \partial y}|_{x=1} ∂ x ∂ y ∂ 2 z ∣ x = 1
【2017-12-10 分】 设函数 f ( u , v ) \displaystyle f(u, v) f ( u , v ) y = f ( e x , cos x ) \displaystyle y=f(e^{x}, \cos x) y = f ( e x , cos x ) d y d x ∣ x = 0 , d 2 y d x 2 ∣ x = 0 \displaystyle \dfrac{d y}{d x}|_{x=0},\dfrac{d^{2} y}{d x^{2}}|_{x=0} d x d y ∣ x = 0 , d x 2 d 2 y ∣ x = 0
【2019-3-10 分】 设函数 f ( u , v ) \displaystyle f(u, v) f ( u , v ) g ( x , y ) = x y − f ( x + y , x − y ) \displaystyle g(x, y)=x y-f(x+y,x-y) g ( x , y ) = x y − f ( x + y , x − y ) ∂ g ∂ x 2 + ∂ g ∂ x ∂ y + ∂ g ∂ y 2 \displaystyle \dfrac{\partial g}{\partial x^{2}} + \dfrac{\partial g}{\partial x \partial y} + \dfrac{\partial g}{\partial y^{2}} ∂ x 2 ∂ g + ∂ x ∂ y ∂ g + ∂ y 2 ∂ g
【2019-2-11 分】 已知函数 u ( x , y ) \displaystyle u(x, y) u ( x , y ) 2 ∂ 2 u ∂ x 2 − 2 ∂ 2 u ∂ y 2 + 3 ∂ u ∂ x + 3 ∂ u ∂ y = 0 \displaystyle 2 \dfrac{\partial^{2} u}{\partial x^{2}}-2 \dfrac{\partial^{2} u}{\partial y^{2}}+3 \dfrac{\partial u}{\partial x}+3 \dfrac{\partial u}{\partial y}=0 2 ∂ x 2 ∂ 2 u − 2 ∂ y 2 ∂ 2 u + 3 ∂ x ∂ u + 3 ∂ y ∂ u = 0 u ( x , y ) = v ( x , y ) e a x + b y \displaystyle u(x, y)=v(x, y) e^{a x+b y} u ( x , y ) = v ( x , y ) e a x + b y v ( x , y ) \displaystyle v(x, y) v ( x , y )
【2024-2-12 分】 已知函数 f ( u , v ) \displaystyle f(u, v) f ( u , v ) g ( x , y ) = f ( 2 x + y , 3 x − y ) \displaystyle g(x, y)=f(2 x+y, 3 x-y) g ( x , y ) = f ( 2 x + y , 3 x − y ) ∂ 2 g ∂ x 2 + ∂ 2 g ∂ x ∂ y − 6 ∂ 2 g ∂ y 2 = 1 \displaystyle \dfrac{\partial^{2} g}{\partial x^{2}}+\dfrac{\partial^{2} g}{\partial x \partial y}-6 \dfrac{\partial^{2} g}{\partial y^{2}}=1 ∂ x 2 ∂ 2 g + ∂ x ∂ y ∂ 2 g − 6 ∂ y 2 ∂ 2 g = 1 ∂ 2 f ∂ u ∂ v \displaystyle \dfrac{\partial^{2} f}{\partial u \partial v} ∂ u ∂ v ∂ 2 f ∂ f ( u , 0 ) ∂ u = u e − u \displaystyle \dfrac{\partial f(u, 0)}{\partial u}=u e^{-u} ∂ u ∂ f ( u , 0 ) = u e − u f ( 0 , v ) = v 2 50 − 1 \displaystyle f(0, v)=\dfrac{v^{2}}{50}-1 f ( 0 , v ) = 50 v 2 − 1 f ( u , v ) \displaystyle f(u, v) f ( u , v )
【2025-1-12 分】 已知函数 f ( u ) \displaystyle f(u) f ( u ) ( 0 , + ∞ ) \displaystyle (0,+\infty) ( 0 , + ∞ ) g ( x , y ) = f ( x y ) \displaystyle g(x, y)=f(\dfrac{x}{y}) g ( x , y ) = f ( y x ) g ( x , y ) \displaystyle g(x, y) g ( x , y ) x 2 ∂ 2 g ∂ x 2 + x y ∂ 2 g ∂ x ∂ y + y 2 ∂ 2 g ∂ y 2 = 1 \displaystyle x^{2} \dfrac{\partial^{2} g}{\partial x^{2}}+x y \dfrac{\partial^{2} g}{\partial x \partial y}+y^{2} \dfrac{\partial^{2} g}{\partial y^{2}}=1 x 2 ∂ x 2 ∂ 2 g + x y ∂ x ∂ y ∂ 2 g + y 2 ∂ y 2 ∂ 2 g = 1 g ( x , x ) = 1 \displaystyle g(x, x)=1 g ( x , x ) = 1 ∂ g ∂ x ∣ ( x , x ) = 2 x \displaystyle \dfrac{\partial g}{\partial x}|_{(x, x)}=\dfrac{2}{x} ∂ x ∂ g ∣ ( x , x ) = x 2 f ( u ) \displaystyle f(u) f ( u )