已知曲线y=1/2x^2+1/2上一点P(1,1),则点P处的切线的倾斜角为,
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![已知曲线y=1/2x^2+1/2上一点P(1,1),则点P处的切线的倾斜角为,](/uploads/image/z/13357936-64-6.jpg?t=%E5%B7%B2%E7%9F%A5%E6%9B%B2%E7%BA%BFy%3D1%2F2x%5E2%2B1%2F2%E4%B8%8A%E4%B8%80%E7%82%B9P%281%2C1%29%2C%E5%88%99%E7%82%B9P%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E7%9A%84%E5%80%BE%E6%96%9C%E8%A7%92%E4%B8%BA%2C)
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已知曲线y=1/2x^2+1/2上一点P(1,1),则点P处的切线的倾斜角为,
已知曲线y=1/2x^2+1/2上一点P(1,1),则点P处的切线的倾斜角为,
已知曲线y=1/2x^2+1/2上一点P(1,1),则点P处的切线的倾斜角为,
求导
y'=1/2*2x=x
x=1,k=y'=1
切线斜率是1
所以倾斜角是arctan1=π/4