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

选项

答案在φ(t)=f[t,f(t,t2)]中令u=t,υ=f(t,t2),得φ(t)=f(u,υ),φ’(t)=f’1(u,υ). [*]=f’1(u,υ).1+f’2(u,υ).[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],所以 φ’(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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