作业帮大一下复合偏导,高数题：设f(x,y)可微且φ(x)=f[x,f(x,f(x,x))],f(1,1)=1,fx(1,1)=a,fy(1,1)=b,则φ'(1)设f(x,y)可微且φ(x)=f[x,f(x,f(x,x))],f(1,1)=1,fx(1,1)=a,fy(1,1)=b,则φ'(1)=?.求具体过程,如果打字不好表达,可以写纸
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![大一下复合偏导,高数题：设f(x,y)可微且φ(x)=f[x,f(x,f(x,x))],f(1,1)=1,fx(1,1)=a,fy(1,1)=b,则φ'(1)设f(x,y)可微且φ(x)=f[x,f(x,f(x,x))],f(1,1)=1,fx(1,1)=a,fy(1,1)=b,则φ'(1)=?.求具体过程,如果打字不好表达,可以写纸](/uploads/image/z/5569364-20-4.jpg?t=%E5%A4%A7%E4%B8%80%E4%B8%8B%E5%A4%8D%E5%90%88%E5%81%8F%E5%AF%BC%2C%E9%AB%98%E6%95%B0%E9%A2%98%EF%BC%9A%E8%AE%BEf%28x%2Cy%29%E5%8F%AF%E5%BE%AE%E4%B8%94%CF%86%28x%29%3Df%5Bx%2Cf%28x%2Cf%28x%2Cx%29%29%5D%2Cf%281%2C1%29%3D1%2Cfx%281%2C1%29%3Da%2Cfy%281%2C1%29%3Db%2C%E5%88%99%CF%86%27%281%29%E8%AE%BEf%28x%2Cy%29%E5%8F%AF%E5%BE%AE%E4%B8%94%CF%86%28x%29%3Df%5Bx%2Cf%28x%2Cf%28x%2Cx%29%29%5D%2Cf%281%2C1%29%3D1%2Cfx%281%2C1%29%3Da%2Cfy%281%2C1%29%3Db%2C%E5%88%99%CF%86%27%281%29%3D%3F.%E6%B1%82%E5%85%B7%E4%BD%93%E8%BF%87%E7%A8%8B%2C%E5%A6%82%E6%9E%9C%E6%89%93%E5%AD%97%E4%B8%8D%E5%A5%BD%E8%A1%A8%E8%BE%BE%2C%E5%8F%AF%E4%BB%A5%E5%86%99%E7%BA%B8)
大一下复合偏导,高数题：设f(x,y)可微且φ(x)=f[x,f(x,f(x,x))],f(1,1)=1,fx(1,1)=a,fy(1,1)=b,则φ'(1)设f(x,y)可微且φ(x)=f[x,f(x,f(x,x))],f(1,1)=1,fx(1,1)=a,fy(1,1)=b,则φ'(1)=?.求具体过程,如果打字不好表达,可以写纸
大一下复合偏导,高数题：设f(x,y)可微且φ(x)=f[x,f(x,f(x,x))],f(1,1)=1,fx(1,1)=a,fy(1,1)=b,则φ'(1)
设f(x,y)可微且φ(x)=f[x,f(x,f(x,x))],f(1,1)=1,fx(1,1)=a,fy(1,1)=b,则φ'(1)=?.求具体过程,如果打字不好表达,可以写纸上发图片,或发到yhh321@yeah.net,
大一下复合偏导,高数题：设f(x,y)可微且φ(x)=f[x,f(x,f(x,x))],f(1,1)=1,fx(1,1)=a,fy(1,1)=b,则φ'(1)设f(x,y)可微且φ(x)=f[x,f(x,f(x,x))],f(1,1)=1,fx(1,1)=a,fy(1,1)=b,则φ'(1)=?.求具体过程,如果打字不好表达,可以写纸
令u=x、t=f(x,u)、y=f(x,t),则φ(x)=f(x,y)；φ'(x)=f'x(x,y)＋f'y(x,y)y'①； y'=f'x(x,t)＋f't(x,t)t'②；t'=f'x(x,u)＋f'u(x,u)u'③；u'=1④；所以u(1)=1、t(1)=f(1,1)=1、y(1)=f(1,1)=1,t'(1)=a＋b,y'(1)=a＋b(a＋b),φ'(1)=a＋b[a＋b(a＋b)]=a＋b[a＋ab＋b^2],即φ'(1)=a＋ab＋ab^2＋b^3.解毕.