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Câu hỏi: Tìm đạo hàm của các hàm số sau:
\(y = {x \over {{{\left( {1 - x} \right)}^2}{{\left({1 + x} \right)}^3}}}.\)
\(y = {x \over {{{\left( {1 - x} \right)}^2}{{\left({1 + x} \right)}^3}}}.\)
Lời giải chi tiết
$\begin{array}{l}
{y^\prime } = \dfrac{{{{(x)}^\prime }{{(1 - x)}^2}{{(1 + x)}^3} - x{{\left[ {{{(1 - x)}^2}{{(1 + x)}^3}} \right]}^\prime }}}{{{{\left[ {{{(1 - x)}^2}{{(1 + x)}^3}} \right]}^2}}}\\
= \dfrac{{{{(1 - x)}^2}{{(1 + x)}^3} - x\left\{ {{{\left[ {{{(1 - x)}^2}} \right]}^\prime }{{(1 + x)}^3} + {{(1 - x)}^2}{{\left[ {{{(1 + x)}^3}} \right]}^\prime }} \right\}}}{{{{\left[ {{{(1 - x)}^2}{{(1 + x)}^3}} \right]}^2}}}\\
= \dfrac{{{{(1 - x)}^2}{{(1 + x)}^3} - x\left[ {2(1 - x){{(1 - x)}^\prime }{{(1 + x)}^3} + {{(1 - x)}^2} \cdot 3{{(1 + x)}^2}{{(1 + x)}^\prime }} \right]}}{{{{\left[ {{{(1 - x)}^2}{{(1 + x)}^3}} \right]}^2}}}\\
= \dfrac{{{{(1 - x)}^2}{{(1 + x)}^3} - x\left[ { - 2\left( {1 - x} \right){{\left( {1 + x} \right)}^3} + 3{{\left( {1 - x} \right)}^2}{{\left( {1 + x} \right)}^2}} \right]}}{{{{\left( {1 - x} \right)}^4}{{\left( {1 + x} \right)}^6}}}\\
= \dfrac{{{{(1 - x)}^2}{{(1 + x)}^3} + 2x(1 - x){{(1 + x)}^3} - 3x{{(1 - x)}^2}{{(1 + x)}^2}}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
= \dfrac{{(1 - x)(1 + x) + 2x(1 + x) - 3x(1 - x)}}{{{{(1 - x)}^3}{{(1 + x)}^4}}}\\
= \dfrac{{1 - {x^2} + 2x + 2{x^2} - 3x + 3{x^2}}}{{{{(1 - x)}^3}{{(1 + x)}^4}}}\\
= \dfrac{{1 - x + 4{x^2}}}{{{{(1 - x)}^3}{{(1 + x)}^4}}}
\end{array}$
$\begin{array}{l}
{y^\prime } = \dfrac{{{{(x)}^\prime }{{(1 - x)}^2}{{(1 + x)}^3} - x{{\left[ {{{(1 - x)}^2}{{(1 + x)}^3}} \right]}^\prime }}}{{{{\left[ {{{(1 - x)}^2}{{(1 + x)}^3}} \right]}^2}}}\\
= \dfrac{{{{(1 - x)}^2}{{(1 + x)}^3} - x\left\{ {{{\left[ {{{(1 - x)}^2}} \right]}^\prime }{{(1 + x)}^3} + {{(1 - x)}^2}{{\left[ {{{(1 + x)}^3}} \right]}^\prime }} \right\}}}{{{{\left[ {{{(1 - x)}^2}{{(1 + x)}^3}} \right]}^2}}}\\
= \dfrac{{{{(1 - x)}^2}{{(1 + x)}^3} - x\left[ {2(1 - x){{(1 - x)}^\prime }{{(1 + x)}^3} + {{(1 - x)}^2} \cdot 3{{(1 + x)}^2}{{(1 + x)}^\prime }} \right]}}{{{{\left[ {{{(1 - x)}^2}{{(1 + x)}^3}} \right]}^2}}}\\
= \dfrac{{{{(1 - x)}^2}{{(1 + x)}^3} - x\left[ { - 2\left( {1 - x} \right){{\left( {1 + x} \right)}^3} + 3{{\left( {1 - x} \right)}^2}{{\left( {1 + x} \right)}^2}} \right]}}{{{{\left( {1 - x} \right)}^4}{{\left( {1 + x} \right)}^6}}}\\
= \dfrac{{{{(1 - x)}^2}{{(1 + x)}^3} + 2x(1 - x){{(1 + x)}^3} - 3x{{(1 - x)}^2}{{(1 + x)}^2}}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
= \dfrac{{(1 - x)(1 + x) + 2x(1 + x) - 3x(1 - x)}}{{{{(1 - x)}^3}{{(1 + x)}^4}}}\\
= \dfrac{{1 - {x^2} + 2x + 2{x^2} - 3x + 3{x^2}}}{{{{(1 - x)}^3}{{(1 + x)}^4}}}\\
= \dfrac{{1 - x + 4{x^2}}}{{{{(1 - x)}^3}{{(1 + x)}^4}}}
\end{array}$