1 câu trả lời
$\log_{0.7}\dfrac{x^2-9}{x+5}=\log_{0.7}(x^2-9)-\log_{0.7}(x+5)$
$\Rightarrow y'=\left({\log_{0.7}\dfrac{x^2-9}{x+5}}\right)'$
$=\left[{\log_{0.7}(x^2-9)-\log_{0.7}(x+5)}\right]'$
$=\dfrac{(x^2-9)'}{(x^2-9)\ln 0.7}-\dfrac{(x+5)'}{(x+5)\ln 0.7}$
$=\dfrac{2x}{(x^2-9)\ln 0.7}-\dfrac{1}{(x+5)\ln 0.7}$
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