Obliczyć pochodne a) y= odp: b)s= odp.... Question from @Klaudiiia16

January 2019 1 7 Report

Obliczyć pochodne

a) y= $\frac{ x^{2} }{ \sqrt[3]{ x^{3}+1 } }$ odp: $\frac{x( x^{3}+2) }{ ( \sqrt[3]{ x^{3}+1 }) ^{4} }$

b)s= $\sqrt{ \frac{1- \sqrt{t} }{1+ \sqrt{t} } }$ odp $\frac{-1}{2(1+ \sqrt{t}) \sqrt{t(1-t)} }$

Mathēmatikós A)

$( \sqrt[3]{x^{3}+1})' = [(x^{3}+1)^{ \frac{1}{3}}]' = \frac{1}{3} \cdot (x^{3}+1)^{- \frac{2}{3}} \cdot (x^{3}+1)' = \\ \\ = \frac{1}{3 (x+1)^{ \frac{2}{3}} } \cdot 3x^{2} = \frac{x^{2}}{ \sqrt[3]{x+1}^{2} }$

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$(x^{2})' = 2x$

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$y = \frac{x^{2}}{ \sqrt[3]{x^{3}+1} } = \frac{f(x)}{g(x)} \\ \\ y' = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{g(x)^{2}} = \frac{2x( \sqrt[3]{x^{3}+1}) - x^{2} \cdot \frac{x^{2}}{ ( \sqrt[3]{x^{3}+1})^{2} } }{( \sqrt[3]{x^{3}+1})^{2}} =$

$= \frac{ \frac{2x \cdot ( \sqrt[3]{x^{3}+1} )^{3}}{(\sqrt[3]{x^{3}+1})^{2}} - \frac{x^{4}}{(\sqrt[3]{x^{3}+1})^{2}} }{(\sqrt[3]{x^{3}+1})^{2}} = \frac{ \frac{2x(x^{3}+1) - x^{4}}{(\sqrt[3]{x^{3}+1})^{2}} }{(\sqrt[3]{x^{3}+1})^{2}} = \frac{2x^{4} + 2x - x^{4}}{(\sqrt[3]{x^{3}+1})^{4}} = \\ \\ = \frac{x^{4}+2x}{(\sqrt[3]{x^{3}+1})^{4}} =\boxed{ \frac{x(x^{3}+2)}{(\sqrt[3]{x^{3}+1})^{4}} }$

b)

$\sqrt{ \frac{1- \sqrt{t} }{1+ \sqrt{t} } } = \frac{ \sqrt{1-\sqrt{t}} }{ \sqrt{1+\sqrt{t}} }$

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$( \sqrt{1-\sqrt{t}} )' =[(1- \sqrt{t})^{ \frac{1}{2}}]' = \frac{1}{2} \cdot (1 - \sqrt{t} )^{- \frac{1}{2}}} \cdot (1 - \sqrt{t}) '= \\ \\ = \frac{1}{2(1- \sqrt{t})^{ \frac{1}{2}} } \cdot (- \frac{1}{2 \sqrt{t} } ) = - \frac{1}{4 \sqrt{t}( \sqrt{1-\sqrt{t}} ) }$

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$\sqrt{1+\sqrt{t} })' = [(1+\sqrt{t})^{ \frac{1}{2} }]' = \frac{1}{2} \cdot (1+\sqrt{t})^{- \frac{1}{2}} \cdot (1+\sqrt{t})' = \\ \\ = \frac{1}{2(1+ \sqrt{t})^{ \frac{1}{2}} } \cdot \frac{1}{2 \sqrt{t} }= \frac{1}{4 \sqrt{t}( \sqrt{1+ \sqrt{t} }) }$

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$s = \sqrt{ \frac{1- \sqrt{t} }{1+ \sqrt{t} } } \\ \\ s' = ( \sqrt{ \frac{1- \sqrt{t} }{1+ \sqrt{t} } } )' =(\frac{ \sqrt{1-\sqrt{t}} }{ \sqrt{1+\sqrt{t}} } )' = \frac{( \sqrt{1 - \sqrt{t} })'( \sqrt{1+\sqrt{t}} - ( \sqrt{1-\sqrt{t}})( \sqrt{1+\sqrt{t}})' }{( \sqrt{1+\sqrt{t}})^{2} } =$

$= \frac{ - \frac{1}{4 \sqrt{t}( \sqrt{1-\sqrt{t}} ) } \cdot ( \sqrt{1+ \sqrt{t} )} - \frac{1}{4 \sqrt{t}( \sqrt{1+ \sqrt{t} }) } \cdot ( \sqrt{1 - \sqrt{t} } ) }{1+ \sqrt{t} } = - \frac{ \frac{ \sqrt{1+ \sqrt{t} } }{4 \sqrt{t}( \sqrt{1- \sqrt{t} })} + \frac{ \sqrt{1 - \sqrt{t}} }{4 \sqrt{t}( \sqrt{1+ \sqrt{t} })} }{1+ \sqrt{t} }=$

$= - \frac{ \frac{( \sqrt{1+ \sqrt{t}})( \sqrt{1 + \sqrt{t}}) }{4 \sqrt{t}( \sqrt{1- \sqrt{t} })( \sqrt{1 + \sqrt{t} }) } + \frac{( \sqrt{1- \sqrt{t}})( \sqrt{1- \sqrt{t} } }{4 \sqrt{t}( \sqrt{1+ \sqrt{t} })( \sqrt{1 - \sqrt{t} }) } }{1+ \sqrt{t} } =-\frac{ \frac{( \sqrt{1+ \sqrt{t} })^{2} +( \sqrt{1- \sqrt{t}})^{2} }{4 \sqrt{t}( \sqrt{(1+ \sqrt{t})(1- \sqrt{t})} }}{1+ \sqrt{t} } =$

$= - \frac{ \frac{1+ \sqrt{t}+1 - \sqrt{t} }{4 \sqrt{t}( \sqrt{1 - t)} } }{1+ \sqrt{t} } =- \frac{2}{4 \sqrt{t}( \sqrt{1-t})(1+ \sqrt{t} ) } =\boxed{ - \frac{1}{2 \sqrt{t}( \sqrt{1- t})(1+ \sqrt{t})} } }$

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