求曲线y=sinx/x在点M(π,0)处的切线方程解答:y’=(xcosx-sinx)/x²∵切点M为(π,0)∴切线方程的斜率k=(πcosπ-sinπ)/π²=1/π设切线方程为y=(1/π)x+b,∴0=(1/π)*π+b,即b=-1,∴曲线y=sinx/x
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![求曲线y=sinx/x在点M(π,0)处的切线方程解答:y’=(xcosx-sinx)/x²∵切点M为(π,0)∴切线方程的斜率k=(πcosπ-sinπ)/π²=1/π设切线方程为y=(1/π)x+b,∴0=(1/π)*π+b,即b=-1,∴曲线y=sinx/x](/uploads/image/z/3016486-46-6.jpg?t=%E6%B1%82%E6%9B%B2%E7%BA%BFy%3Dsinx%2Fx%E5%9C%A8%E7%82%B9M%EF%BC%88%CF%80%2C0%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E8%A7%A3%E7%AD%94%EF%BC%9Ay%E2%80%99%3D%EF%BC%88xcosx-sinx%EF%BC%89%2Fx%26sup2%3B%E2%88%B5%E5%88%87%E7%82%B9M%E4%B8%BA%EF%BC%88%CF%80%2C0%EF%BC%89%E2%88%B4%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E7%9A%84%E6%96%9C%E7%8E%87k%3D%EF%BC%88%CF%80cos%CF%80-sin%CF%80%EF%BC%89%2F%CF%80%26sup2%3B%3D1%2F%CF%80%E8%AE%BE%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E4%B8%BAy%3D%EF%BC%881%2F%CF%80%EF%BC%89x%2Bb%2C%E2%88%B40%3D%EF%BC%881%2F%CF%80%EF%BC%89%2A%CF%80%2Bb%2C%E5%8D%B3b%3D-1%2C%E2%88%B4%E6%9B%B2%E7%BA%BFy%3Dsinx%2Fx)
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