Question

arctanx(x²+y²)=ye^√x,求dy/dx

Answer 1

问：arctanx(x²+y²)=ye^√x,求dy/dx

答：您好[爱心],感谢您的耐心等待，根据您的提问【arctanx(x²+y²)=ye^√x,求dy/dx】以为您编辑好了解答回答如下：解析：此题需要求出参数方程的导数。首先，对于式子 arctan(x² + y²) = ye^√x，将两边分别求导。左边：d/dx[arctan(x² + y²)] = d/dx[tan^(-1)(x² + y²)]因为 arctan 的导数为 1 / (1 + x²)，所以：= 1 / (1 + (x² + y²)²) * (2x)= 2x / (1 + (x² + y²)²)右边：d/dx[ye^√x] = (√x * e^√x + y' * e^√x)因为 y = ye^√x，所以：y' = dy/dx = e^√x * de^√x/dx = e^√x * (1 / 2x^(-1/2)) = e^√x * (2√x)^(-1)因此：= (√x * e^√x + e^√x * (2√x)^(-1)) = (√x + 1 / 2x) * e^√x综上所述：2x / (1 + (x² + y²)²) = (√x + 1 / 2x) * e^√x移项后求解 y'，得：y' = (2x * e^√x) / (1 + (x² + y²)²) - (√x + 1 / 2x) * e^√x * (2x) / (1 + (x² + y²)²)= (2x * e^√x - (√x + 1 / 2x) * 2x * e^√x) / (1 + (x² + y²)²)= (-(√x + 1 / 2x) * 2x * e^√x) / (1 + (x² + y²)²)因此，dy/dx = -(√x + 1 / 2x) * 2x * e^√x / (1 + (x² + y²)²)。