найти dy/dx [tex] \left \{ {{x=lny} \atop {y=} \frac{1}{} t^{2}-5 } \right. [/tex]. Created by mkineshma. algebra-ru

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mkineshma @mkineshma

July 2022 1 2 Report

найти dy/dx
[tex] \left \{ {{x=lny} \atop {y=} \frac{1}{} t^{2}-5 } \right. [/tex]

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$\left \{ {{x=ln t} \atop {y=\frac{1}{t^2-5}}} \right. \\\\x'_{t}=\frac{1}{t}\\\\y'_{t}=- \frac{2t}{(t^2-5)^2} \\\\y'_{x}= \frac{dy}{dx}= \frac{y'_{t}}{x'_{t}} =- \frac{2t^2}{(t^2-5)^2}$

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