Question

find du/dt. If u= xyz ,x= e^-t ,y=e^-t sin^2t ,z=sint​

Answer 1

$\rm u = xyz$

$\rm \implies u =( {e}^{ - t} ) \{ {e}^{ - t} { \sin}^{2}( t) \} \{\sin(t) \}$

$\rm \implies u = {e}^{ -2 t} { \sin}^{3}( t)$

Now, differentiation w.r.t 't':

$\rm \implies \dfrac{du}{dt} = \dfrac{d \bigg \{{e}^{ -2 t} { \sin}^{3}( t) \bigg \}}{dt}$

$\rm \implies \dfrac{du}{dt} ={ \sin}^{3}( t) \dfrac{d \bigg \{{e}^{ -2 t} \bigg \}}{dt} + {e}^{ - 2t} \dfrac{d \bigg \{{ \sin}^{ 3} (t) \bigg \}}{dt}$

$\rm \implies \dfrac{du}{dt} = - 2{ \sin}^{3}( t) {e}^{ -2 t} + 2 { \sin}^{2} (t) \cos(t) {e}^{ - 2t}$

$\rm \implies \dfrac{du}{dt} = {e}^{ -2 t} \bigg \{ 2 { \sin}^{2} (t) \cos(t) - 2{ \sin}^{3}( t) \bigg \}$

$\boxed{ \rm \implies \dfrac{du}{dt} = 2{ \sin}^{2}( t) {e}^{ -2 t} \bigg \{ \cos(t) - \sin( t) \bigg \}}$

Hope It Helps.