求y=(x/4)^2+[(100-x)/4]^2的导数正确过程是y=(x/4)^2+[(100-x)/4]^2=(2x^2-200x+10000)/16 y'=(4x-200)/16=0 x=50我想问的是为什么不能这么算y'=[(x/4)^2]'+{[(100-x)/4]^2}'=x/2+(100-x)/2=0 这样算x就无解了,这种算法哪里错
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![求y=(x/4)^2+[(100-x)/4]^2的导数正确过程是y=(x/4)^2+[(100-x)/4]^2=(2x^2-200x+10000)/16 y'=(4x-200)/16=0 x=50我想问的是为什么不能这么算y'=[(x/4)^2]'+{[(100-x)/4]^2}'=x/2+(100-x)/2=0 这样算x就无解了,这种算法哪里错](/uploads/image/z/5349284-44-4.jpg?t=%E6%B1%82y%3D%28x%2F4%29%5E2%2B%5B%28100-x%29%2F4%5D%5E2%E7%9A%84%E5%AF%BC%E6%95%B0%E6%AD%A3%E7%A1%AE%E8%BF%87%E7%A8%8B%E6%98%AFy%3D%28x%2F4%29%5E2%2B%5B%28100-x%29%2F4%5D%5E2%3D%EF%BC%882x%5E2-200x%2B10000%29%2F16+y%27%3D%284x-200%29%2F16%3D0+x%3D50%E6%88%91%E6%83%B3%E9%97%AE%E7%9A%84%E6%98%AF%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%8D%E8%83%BD%E8%BF%99%E4%B9%88%E7%AE%97y%27%3D%5B%28x%2F4%29%5E2%5D%27%2B%7B%5B%28100-x%29%2F4%5D%5E2%7D%27%3Dx%2F2%2B%28100-x%29%2F2%3D0+%E8%BF%99%E6%A0%B7%E7%AE%97x%E5%B0%B1%E6%97%A0%E8%A7%A3%E4%BA%86%2C%E8%BF%99%E7%A7%8D%E7%AE%97%E6%B3%95%E5%93%AA%E9%87%8C%E9%94%99)
求y=(x/4)^2+[(100-x)/4]^2的导数正确过程是y=(x/4)^2+[(100-x)/4]^2=(2x^2-200x+10000)/16 y'=(4x-200)/16=0 x=50我想问的是为什么不能这么算y'=[(x/4)^2]'+{[(100-x)/4]^2}'=x/2+(100-x)/2=0 这样算x就无解了,这种算法哪里错
求y=(x/4)^2+[(100-x)/4]^2的导数
正确过程是y=(x/4)^2+[(100-x)/4]^2=(2x^2-200x+10000)/16 y'=(4x-200)/16=0 x=50
我想问的是为什么不能这么算y'=[(x/4)^2]'+{[(100-x)/4]^2}'=x/2+(100-x)/2=0 这样算x就无解了,这种算法哪里错了?
求y=(x/4)^2+[(100-x)/4]^2的导数正确过程是y=(x/4)^2+[(100-x)/4]^2=(2x^2-200x+10000)/16 y'=(4x-200)/16=0 x=50我想问的是为什么不能这么算y'=[(x/4)^2]'+{[(100-x)/4]^2}'=x/2+(100-x)/2=0 这样算x就无解了,这种算法哪里错
你求导数的方法错了=x/8-(100-x)/8
比如{[(100-x)/4]^2}'=2*(100-x)/4*(-1/4)
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