Question

Найти полную производную

Answer 1

$z=e^{ \frac{ty}{x^2} }; x=ctgt;y=t^5 \\ \frac{dz}{dt} = \frac{dz}{dx} \frac{dx}{dt} +\frac{dz}{dy} \frac{dy}{dt} \\\\ \frac{dz}{dx} =e^{ \frac{ty}{x^2} }*(\frac{ty}{x^2} )'_x=e^{ \frac{ty}{x^2} }*(tyx^{-2} )'_x=e^{ \frac{ty}{x^2} }*(-2tyx^{-3} )= - \frac{2ty}{x^3} e^{ \frac{ty}{x^2} } \\ \frac{dz}{dy} =e^{ \frac{ty}{x^2} }*(\frac{ty}{x^2} )'_y=e^{ \frac{ty}{x^2} }*\frac{t}{x^2} =\frac{t}{x^2}e^{ \frac{ty}{x^2} } \\ \frac{dx}{dt} =- \frac{1}{sin^2t} \\ \frac{dy}{dt} =5t^4$

$\frac{dz}{dt} = - \frac{2ty}{x^3} e^{ \frac{ty}{x^2} }*(- \frac{1}{sin^2t} )+\frac{t}{x^2}e^{ \frac{ty}{x^2} }*5t^4= \\ =\frac{2ty}{x^3sin^2t} e^{\frac{ty}{x^2} }+\frac{5t^5}{x^2}e^{ \frac{ty}{x^2}} = (\frac{2y}{xsin^2t} +5t^4) \frac{t}{x^2} e^{ \frac{ty}{x^2}} = \\ =(\frac{2t^5}{ctgtsin^2t} +5t^4) \frac{t}{ctg^2t} e^{ \frac{t^6}{ctg^2t}} = \\\ =\boxed{(\frac{2t}{costsint} +5) \frac{t^5}{ctg^2t} e^{ \frac{t^6}{ctg^2t}} }$