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Câu hỏi: Đối với hàm số $y=\ln \dfrac{1}{x+1}$, khẳng định nào sau đây là khẳng định đúng?
A. $x{y}'-1=-{{e}^{y}}.$
B. $x{y}'+1=-{{e}^{y}}.$
C. $x{y}'-1={{e}^{y}}.$
D. $x{y}'+1={{e}^{y}}.$
A. $x{y}'-1=-{{e}^{y}}.$
B. $x{y}'+1=-{{e}^{y}}.$
C. $x{y}'-1={{e}^{y}}.$
D. $x{y}'+1={{e}^{y}}.$
Ta có: $y=\ln \dfrac{1}{x+1}=\ln 1-\ln \left( x+1 \right)=-\ln \left( x+1 \right)$
Suy ra ${y}'={{\left[ -\ln \left( x+1 \right) \right]}^{\prime }}=-\dfrac{{{\left( x+1 \right)}^{\prime }}}{x+1}=-\dfrac{1}{x+1}$
Do đó $x{y}'-1=x.\left( -\dfrac{1}{x+1} \right)-1=\dfrac{-x-\left( x+1 \right)}{x+1}=\dfrac{-2x-1}{x+1}$
$x{y}'+1=x.\left( -\dfrac{1}{x+1} \right)+1=\dfrac{-x+\left( x+1 \right)}{x+1}=\dfrac{1}{x+1}$
${{e}^{y}}={{e}^{-\ln \left( x+1 \right)}}={{e}^{\ln \dfrac{1}{x+1}}}=\dfrac{1}{x+1}=x{y}'+1$
Vậy $x{y}'+1={{e}^{y}}.$
Suy ra ${y}'={{\left[ -\ln \left( x+1 \right) \right]}^{\prime }}=-\dfrac{{{\left( x+1 \right)}^{\prime }}}{x+1}=-\dfrac{1}{x+1}$
Do đó $x{y}'-1=x.\left( -\dfrac{1}{x+1} \right)-1=\dfrac{-x-\left( x+1 \right)}{x+1}=\dfrac{-2x-1}{x+1}$
$x{y}'+1=x.\left( -\dfrac{1}{x+1} \right)+1=\dfrac{-x+\left( x+1 \right)}{x+1}=\dfrac{1}{x+1}$
${{e}^{y}}={{e}^{-\ln \left( x+1 \right)}}={{e}^{\ln \dfrac{1}{x+1}}}=\dfrac{1}{x+1}=x{y}'+1$
Vậy $x{y}'+1={{e}^{y}}.$
Đáp án D.