♤三4求一道高数题
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u=f(t)，t=e^x+e^(-y)
∂u/∂x=f'(t)*∂t/∂x=f'(t)*e^x
∂u/∂y=f'(t)*∂t/∂y=f'(t)*[-e^(-y)]=-f'(t)*e^(-y)
∂²u/∂x²=f''(t)*∂t/∂x*e^x+f'(t)*e^x=f''(t)*e^(2x)+f'(t)*e^x
∂²u/∂y²=-f''(t)*∂t/∂y*e^(-y)+f'(t)*e^(-y)=f''(t)*e^(-2y)+f'(t)*e^(-y)
所以∂²u/∂x²-∂²u/∂y²
=f''(t)*e^(2x)+f'(t)*e^x-f''(t)*e^(-2y)-f'(t)*e^(-y)
=f''(t)*[e^(2x)-e^(-2y)]+f'(t)*[e^x-e^(-y)]
∂u/∂x=f'(t)*∂t/∂x=f'(t)*e^x
∂u/∂y=f'(t)*∂t/∂y=f'(t)*[-e^(-y)]=-f'(t)*e^(-y)
∂²u/∂x²=f''(t)*∂t/∂x*e^x+f'(t)*e^x=f''(t)*e^(2x)+f'(t)*e^x
∂²u/∂y²=-f''(t)*∂t/∂y*e^(-y)+f'(t)*e^(-y)=f''(t)*e^(-2y)+f'(t)*e^(-y)
所以∂²u/∂x²-∂²u/∂y²
=f''(t)*e^(2x)+f'(t)*e^x-f''(t)*e^(-2y)-f'(t)*e^(-y)
=f''(t)*[e^(2x)-e^(-2y)]+f'(t)*[e^x-e^(-y)]
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