Mathematica求偏导还不会？赶紧进来抄作业

　　分享兴趣、传播快乐、增长见闻、留下美好，大家好，这里是LearningYard学苑，今天小编为大家带来文章：用mathematica求偏导
　　Share interest, spread happiness, increase knowledge, and stay good. Hello, everyone. This is LearningYard Academy. Today I will bring you an article: Using mathematica to ask for partial guidance
　　大家下午好呀
　　上一次给大家介绍了用  Mathematica作三维图的方法
　　附上补课链接高颜值的三维图片，建议学习收藏
　　今天小编分享的也是Mathematica哦
　　老规矩，感兴趣的小伙伴 赶紧@你的小伙伴一起学习吧！
　　Good afternoon everyone
　　Last time, I introduced you to the method of using Mathematica to make 3D graphs.
　　Attach a link to make-up lessons高颜值的三维图片，建议学习收藏
　　Today the editor is sharing Mathematica too
　　Old rules, interested friends, hurry up @you friends to learn together!
　　接下来开始今天的教程学习：  用Mathematica求偏导数与多元函数的极值
　　（有条件的同学建议先复习一下大学课程微积分里的相关内容）
　　Next, start today’s tutorial study: Use Mathematica to find the extreme value of partial derivatives and multivariate functions
　　(Students with conditions suggest to review the relevant content in the college course calculus first)
　　注：在Mathematica中，还是有些微分函数用于直接计算的，如下表所示
　　Note: In Mathematica, there are still some differential functions used for direct calculation, as shown in the following table
　　命令In[1]:= D[Log[x+Sqrt[x^2+y^2] ],x];Simplify[%]
　　通常Mathematica不自动化简微分结果，要借助于Simplify函数
　　Usually Mathematica does not automate the result of simple differentiation, it has to resort to the Simplify function
　　下面开始进阶学习
　　Let＂s start advanced learning
　　小提示：面对复杂的函数时，要先定义函数，再求偏导哦
　　Tips: When facing complex functions, define the function first, and then find the partial derivative.
　　In[1]:=  Clear[z,x,y];z[x,y]:=x^3*Sin[y]+y^3Sin[x]; D[z[x,y],y]
　　/*定义二元函Define a binary function*/
　　Out[1]=
　　In[2]:=  D[z[x,y],y]/.{x->1,y->1} /* 给函数的变量赋值Assign a value to the variable of the functio  */
　　Out[2]=
　　In[3]:=  D[z[x,y],{x,2}]
　　Out[3]=
　　In[4]:=  D[z[x,y],{x,3},{y,3}]
　　Out[4]=
　　继续进阶，求由下列方程所确定的隐函数和导数或偏导数
　　Continue to advance to find the implicit function and derivative or partial derivative determined by the following equations
　　小提示：需要利用solve函数
　　Tips: need to use the solve function
　　今日的第二个主题：求极值
　　Today’s second topic: seeking extreme values
　　In[1]:= f[x_,y_]:=x^2+y^2-12x+16y; t=Solve[{D[f[x,y]= =0,x],D[f[x,y]= =0,y]},{x,y}]
　　Out[1]= {{x->6,y->-8}} (* 驻点 *)
　　In[2]:=  x^2+y^2-25/.t[[1]]
　　Out[2]=75
　　该驻点在圆外，圆内无驻点，故不取极值。下面考虑圆
　　上的最值。这是在约束条件
　　下的条件极值,用Lagrange乘数法求解
　　The stagnation point is outside the circle, and there is no stagnation point inside the circle, so the extreme value is not taken. Next consider the highest value on
　　. This is the conditional extreme value under the constraint bar, which is solved by the Lagrange multiplier method.
　　In[3]:=  Clear[x,y,F,t];F[x_,y_,t_]:=f[x,y]+t(x^2+y^2-25);
　　s=Solve[{D[f[x,y,t]==0,D[f[x,y,t]==0,y],D[F[x,y,t]= =0,t]],{x,y,t}}
　　Out[3]={{t->-3,x->-3,y->4},{t->1,x->3,y->-4}}
　　In[4]:=  F[x,y]/.s[[1]]
　　Out[4]=25
　　In[5]:=  F[x,y]/.s[[2]]
　　Out[5]=-75
　　今天的分享就到这里啦
　　That＂s it for today＂s sharing
　　散会
　　Let＂s adjourn
　　参考资料：网络
　　翻译：Google翻译
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