Question

find du/dt if u=x/y wherex=e^t, Y=logt

Answer 1

Solution :

$\sf{Given, \: the \: function \: of \: u \: is \:} \\ \sf{u = \dfrac{x}{y}\:, where\: \underline{\sf{x = e^{t}}} \: and \:\underline{\sf{y = In(t)}}} \\ \\ \textsf{Now, by substituting the value of} \\ \textsf{x and y in the function of u, we get:} \\ \\ \:\:\:\:\:\: :\implies \sf{u = \dfrac{x}{y}} \\ \\ \:\:\:\:\:\: :\implies \sf{u = \dfrac{e^{t}}{In(t)}} \\ \\ \textsf{Hence,the function of u is} \:\: \sf{\dfrac{e^{t}}{In(t)}}$

$\textsf{Now by differentiating the function of u with respect to t}$ $\textsf{by using the quotient rule of differentiation, we get :}$

$\underline{\sf{Quotient\:rule\: of\:differentiation :-}} \\ \\ \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{(v)\dfrac{d(u)}{dx} - (u)\dfrac{d(v)}{dx}}{v^{2}}}$

$:\implies \sf{\dfrac{d}{dt}\bigg[\dfrac{e^{t}}{In(t)}\bigg] = \dfrac{[In(t)]\dfrac{d(e^{t})}{dt} - (e^{t})\dfrac{d[In(t)]}{dt}}{[In(t)]^{2}}}$

$\underline{\sf{We\:know\:that\:the\:differentiation\:of\::-}} \\ \\ \bullet \sf{\dfrac{d(e^{x})}{dx} = e^{x}} \\ \\ \bullet \sf{\dfrac{d[In(x)]}{dx} = \dfrac{1}{x}} \\ \\ \textsf{So, by substituting them in the equation, we get :}$

$:\implies \sf{\dfrac{d}{dt}\bigg[\dfrac{e^{t}}{In(t)}\bigg] = \dfrac{In(t) \times e^{t} - (e^{t}) \times \dfrac{1}{t}}{In^{2}(t)}} \\ \\ \\ :\implies \sf{\dfrac{d}{dt}\bigg[\dfrac{e^{t}}{In(t)}\bigg] = \dfrac{e^{t}In(t) - \dfrac{e^{t}}{t}}{In^{2}(t)}} \\ \\ \\ \boxed{\therefore \sf{\dfrac{d}{dt}\bigg[\dfrac{e^{t}}{In(t)}\bigg] = \dfrac{e^{t}In(t) - \dfrac{e^{t}}{t}}{In^{2}(t)}}}$