Google
Câu hỏi: Rút gọn:
\(f\left( x \right) = \left({{{x - 1} \over {2\left( {\sqrt x + 1} \right)}} + 1} \right).{2 \over {\sqrt x + 1}}\) \(:{\left( {{{\sqrt {x - 2} } \over {\sqrt {x + 2} + \sqrt {x - 2} }} + {{x - 2} \over {\sqrt {{x^2} - 4} - x + 2}}} \right)^2}\) và tìm f'(x)
\(f\left( x \right) = \left({{{x - 1} \over {2\left( {\sqrt x + 1} \right)}} + 1} \right).{2 \over {\sqrt x + 1}}\) \(:{\left( {{{\sqrt {x - 2} } \over {\sqrt {x + 2} + \sqrt {x - 2} }} + {{x - 2} \over {\sqrt {{x^2} - 4} - x + 2}}} \right)^2}\) và tìm f'(x)
Phương pháp giải
Rút gọn \(f(x)\) rồi tính đạo hàm.
Lời giải chi tiết
\(\begin{array}{l}f\left( x \right) = \left({\frac{{x - 1}}{{2\left( {\sqrt x + 1} \right)}} + 1} \right).\frac{2}{{\sqrt x + 1}}\\:{\left({\frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} + \sqrt {x - 2} }} + \frac{{x - 2}}{{\sqrt {{x^2} - 4} - x + 2}}} \right)^2}\\ = \left({\frac{{\left( {\sqrt x - 1} \right)\left({\sqrt x + 1} \right)}}{{2\left({\sqrt x + 1} \right)}} + 1} \right).\frac{2}{{\sqrt x + 1}}\\:{\left({\frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} + \sqrt {x - 2} }} + \frac{{x - 2}}{{\sqrt {x - 2} \left( {\sqrt {x + 2} - \sqrt {x - 2} } \right)}}} \right)^2}\\ = \left({\frac{{\sqrt x - 1}}{2} + 1} \right).\frac{2}{{\sqrt x + 1}}\\:{\left({\frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} + \sqrt {x - 2} }} + \frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} - \sqrt {x - 2} }}} \right)^2}\\ = \frac{{\sqrt x - 1 + 2}}{2}.\frac{2}{{\sqrt x + 1}}\\:\left({x - 2} \right){\left({\frac{{\sqrt {x + 2} - \sqrt {x - 2} + \sqrt {x + 2} + \sqrt {x - 2} }}{{\left( {\sqrt {x + 2} + \sqrt {x - 2} } \right)\left({\sqrt {x + 2} - \sqrt {x - 2} } \right)}}} \right)^2}\\ = \frac{{\sqrt x + 1}}{2}.\frac{2}{{\sqrt x + 1}}:\left[ {\left({x - 2} \right).{{\left({\frac{{2\sqrt {x + 2} }}{{x + 2 - x + 2}}} \right)}^2}} \right]\\ = 1:\left[ {\left({x - 2} \right).{{\left({\frac{{\sqrt {x + 2} }}{2}} \right)}^2}} \right]\\ = 1:\frac{{\left({x - 2} \right)\left({x + 2} \right)}}{4}\\ = \frac{4}{{{x^2} - 4}}\end{array}\)
\(\Rightarrow f'\left( x \right) = \frac{{ - 4\left({{x^2} - 4} \right)'}}{{{{\left({{x^2} - 4} \right)}^2}}}\)\(= \frac{{ - 4.2x}}{{{{\left( {{x^2} - 4} \right)}^2}}} = - \frac{{8x}}{{{{\left({{x^2} - 4} \right)}^2}}}\)
Rút gọn \(f(x)\) rồi tính đạo hàm.
Lời giải chi tiết
\(\begin{array}{l}f\left( x \right) = \left({\frac{{x - 1}}{{2\left( {\sqrt x + 1} \right)}} + 1} \right).\frac{2}{{\sqrt x + 1}}\\:{\left({\frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} + \sqrt {x - 2} }} + \frac{{x - 2}}{{\sqrt {{x^2} - 4} - x + 2}}} \right)^2}\\ = \left({\frac{{\left( {\sqrt x - 1} \right)\left({\sqrt x + 1} \right)}}{{2\left({\sqrt x + 1} \right)}} + 1} \right).\frac{2}{{\sqrt x + 1}}\\:{\left({\frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} + \sqrt {x - 2} }} + \frac{{x - 2}}{{\sqrt {x - 2} \left( {\sqrt {x + 2} - \sqrt {x - 2} } \right)}}} \right)^2}\\ = \left({\frac{{\sqrt x - 1}}{2} + 1} \right).\frac{2}{{\sqrt x + 1}}\\:{\left({\frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} + \sqrt {x - 2} }} + \frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} - \sqrt {x - 2} }}} \right)^2}\\ = \frac{{\sqrt x - 1 + 2}}{2}.\frac{2}{{\sqrt x + 1}}\\:\left({x - 2} \right){\left({\frac{{\sqrt {x + 2} - \sqrt {x - 2} + \sqrt {x + 2} + \sqrt {x - 2} }}{{\left( {\sqrt {x + 2} + \sqrt {x - 2} } \right)\left({\sqrt {x + 2} - \sqrt {x - 2} } \right)}}} \right)^2}\\ = \frac{{\sqrt x + 1}}{2}.\frac{2}{{\sqrt x + 1}}:\left[ {\left({x - 2} \right).{{\left({\frac{{2\sqrt {x + 2} }}{{x + 2 - x + 2}}} \right)}^2}} \right]\\ = 1:\left[ {\left({x - 2} \right).{{\left({\frac{{\sqrt {x + 2} }}{2}} \right)}^2}} \right]\\ = 1:\frac{{\left({x - 2} \right)\left({x + 2} \right)}}{4}\\ = \frac{4}{{{x^2} - 4}}\end{array}\)
\(\Rightarrow f'\left( x \right) = \frac{{ - 4\left({{x^2} - 4} \right)'}}{{{{\left({{x^2} - 4} \right)}^2}}}\)\(= \frac{{ - 4.2x}}{{{{\left( {{x^2} - 4} \right)}^2}}} = - \frac{{8x}}{{{{\left({{x^2} - 4} \right)}^2}}}\)