lg[x-sqr(x^2-a^2)]的导数怎么求?要过程
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lg[x-sqr(x^2-a^2)]的导数怎么求?要过程
RT
lgX'=(1/X)*lge 不是吗?
RT
lgX'=(1/X)*lge 不是吗?
{lg[x-sqr(x^2-a^2)]}'
=1/[x-sqr(x^2-a^2)*ln10] * [x-sqr(x^2-a^2)]'
=1/[x-sqr(x^2-a^2)*ln10] *{1-[(x^2-a^2)^(1/2)]'}
=1/[x-sqr(x^2-a^2)*ln10] *[1-1/2*(x^2-a^2)^(-1/2)*(x^2-a^2)']
=1/[x-sqr(x^2-a^2)*ln10] *[1-1/2√(x^2-a^2) * 2x]
=1/[x-sqr(x^2-a^2)*ln10] * [1-x/√(x^2-a^2)]
=1/[x-sqr(x^2-a^2)*ln10] * [1-x√(x^2-a^2)/(x^2-a^2)]
=1/[x-sqr(x^2-a^2)*ln10] * [x-sqr(x^2-a^2)]'
=1/[x-sqr(x^2-a^2)*ln10] *{1-[(x^2-a^2)^(1/2)]'}
=1/[x-sqr(x^2-a^2)*ln10] *[1-1/2*(x^2-a^2)^(-1/2)*(x^2-a^2)']
=1/[x-sqr(x^2-a^2)*ln10] *[1-1/2√(x^2-a^2) * 2x]
=1/[x-sqr(x^2-a^2)*ln10] * [1-x/√(x^2-a^2)]
=1/[x-sqr(x^2-a^2)*ln10] * [1-x√(x^2-a^2)/(x^2-a^2)]
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