设函数y=y(x)由方程y=f(x^2+y^2)+f(x+y)确定,且y(0)=2,f(x)是可导函数,f'(2)=1/2,f'(4)=1,则f'(0)的值
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![设函数y=y(x)由方程y=f(x^2+y^2)+f(x+y)确定,且y(0)=2,f(x)是可导函数,f'(2)=1/2,f'(4)=1,则f'(0)的值](/uploads/image/z/9485235-27-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0y%3Dy%28x%29%E7%94%B1%E6%96%B9%E7%A8%8By%3Df%28x%5E2%2By%5E2%29%2Bf%28x%2By%29%E7%A1%AE%E5%AE%9A%2C%E4%B8%94y%280%29%3D2%2Cf%28x%29%E6%98%AF%E5%8F%AF%E5%AF%BC%E5%87%BD%E6%95%B0%2Cf%27%282%29%3D1%2F2%2Cf%27%284%29%3D1%2C%E5%88%99f%27%280%29%E7%9A%84%E5%80%BC)
设函数y=y(x)由方程y=f(x^2+y^2)+f(x+y)确定,且y(0)=2,f(x)是可导函数,f'(2)=1/2,f'(4)=1,则f'(0)的值
设函数y=y(x)由方程y=f(x^2+y^2)+f(x+y)确定,且y(0)=2,f(x)是可导函数,f'(2)=1/2,f'(4)=1,则f'(0)的值
设函数y=y(x)由方程y=f(x^2+y^2)+f(x+y)确定,且y(0)=2,f(x)是可导函数,f'(2)=1/2,f'(4)=1,则f'(0)的值
y=f(x²+y²)+f(x+y)
y'=f'(x²+y²)×(x²+y²)'+f'(x+y)×(x+y)'
=(2x+2yy')f'(x²+y²)+(1+y')f'(x+y)
当x=0时,y=2,那么y'=(0+4y')f'(4)+(1+y')f'(2)
而f'(4)=1,f'(2)=1/2,所以y'=4y'×1+(1+y')×(1/2)
即:y'=4y'+1/2+y'/2,所以y'=-1/7,即f'(0)=-1/7