设z=f(x,y)=arctanx/y ,y=√(x^2+1) ,求dz/dx

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设z=f(x,y)=arctanx/y,y=√(x^2+1),求dz/dx设z=f(x,y)=arctanx/y,y=√(x^2+1),求dz/dx设z=f(x,y)=arctanx/y,y=√(x^

设z=f(x,y)=arctanx/y ,y=√(x^2+1) ,求dz/dx
设z=f(x,y)=arctanx/y ,y=√(x^2+1) ,求dz/dx

设z=f(x,y)=arctanx/y ,y=√(x^2+1) ,求dz/dx
z'=1/[1+(x/y)²]* (x/y)'
=1/[1+(x/y)²] *(y-xy')/y²
=(y-xy')/(y²+x²)
而y'=1/[2√(x²+1)]*2x=x/√(x²+1)
所以z'=[√(x²+1)-x²/√(x²+1)]/(x²+1+x²)=1/[(2x²+1)√(x²+1)]