设函数f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,其中θ∈[0,5π/12],则导数f'(1)其中这一步 因为(f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ, 所以f'(x))=3(sinθ/3)x^2+2(√3cosθ/2)x =(sinθ)x^2+(√3cosθ)xtanθ求导为1/cosθ^2
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![设函数f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,其中θ∈[0,5π/12],则导数f'(1)其中这一步 因为(f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ, 所以f'(x))=3(sinθ/3)x^2+2(√3cosθ/2)x =(sinθ)x^2+(√3cosθ)xtanθ求导为1/cosθ^2](/uploads/image/z/1698627-3-7.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D%28sin%CE%B8%2F3%29x%5E3%2B%28%E2%88%9A3cos%CE%B8%2F2%29x%5E2%2Btan%CE%B8%2C%E5%85%B6%E4%B8%AD%CE%B8%E2%88%88%5B0%2C5%CF%80%2F12%5D%2C%E5%88%99%E5%AF%BC%E6%95%B0f%27%281%29%E5%85%B6%E4%B8%AD%E8%BF%99%E4%B8%80%E6%AD%A5++%E5%9B%A0%E4%B8%BA%28f%28x%29%3D%28sin%CE%B8%2F3%29x%5E3%2B%28%E2%88%9A3cos%CE%B8%2F2%29x%5E2%2Btan%CE%B8%2C++++%E6%89%80%E4%BB%A5f%27%28x%29%29%3D3%28sin%CE%B8%2F3%29x%5E2%2B2%28%E2%88%9A3cos%CE%B8%2F2%29x++++++%3D%28sin%CE%B8%29x%5E2%2B%28%E2%88%9A3cos%CE%B8%29xtan%CE%B8%E6%B1%82%E5%AF%BC%E4%B8%BA1%2Fcos%CE%B8%5E2)
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设函数f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,其中θ∈[0,5π/12],则导数f'(1)其中这一步 因为(f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ, 所以f'(x))=3(sinθ/3)x^2+2(√3cosθ/2)x =(sinθ)x^2+(√3cosθ)xtanθ求导为1/cosθ^2
设函数f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,其中θ∈[0,5π/12],则导数f'(1)
其中这一步 因为(f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,
所以f'(x))=3(sinθ/3)x^2+2(√3cosθ/2)x
=(sinθ)x^2+(√3cosθ)x
tanθ求导为1/cosθ^2 到哪里去了?
设函数f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,其中θ∈[0,5π/12],则导数f'(1)其中这一步 因为(f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ, 所以f'(x))=3(sinθ/3)x^2+2(√3cosθ/2)x =(sinθ)x^2+(√3cosθ)xtanθ求导为1/cosθ^2
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