设函数f(x,y)可微,又f(0,0)=0,f’x(0,0)=a,f’y(0,0)=b,且ψ(t)=f[t,f(t,t2)],求ψ’(0).

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问题 设函数f(x,y)可微,又f(0,0)=0,f’x(0,0)=a,f’y(0,0)=b,且ψ(t)=f[t,f(t,t2)],求ψ’(0).

选项

答案在ψ(t)=f[t,f(t,t2)]中令u=t,v=f(t,t2),得ψ(t)=f(u,v), =f’1(u,v)·1+f’2(u,v)·[f’1(t,t2)·1+f’2(t,t2)·2t] =f’1[t,f(t,t2)]+f’2[t,f(t,t2)]·[f’1(t,t2)+f’2(t,t2)·2t], 又f(0,0)=0,f’x(0,0)=a,f’y(0,0)=b,所以 ψ’(0)=f’1(0,0)+f’2(0,0)·[f’1(0,0)+f’2(0,0)×2×0] =a+b(a+0)=a(1+b).

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