# Find $\large \frac{dy}{dx}$ of the function expressed in parametric form $x=\large\frac{1+log t}{t^2},$$y=\large\frac{3+2log t}{t} ## 1 Answer Toolbox: • To find \large\frac{dy}{dx} in the case of parametric functions,if x=\phi(t) and y=\psi(t),then \large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
Step 1:
Given $x=\large\frac{1+\log t}{t^2}$
$y=\large\frac{3+2\log t}{t}$
Consider $x=\large\frac{1+\log t}{t^2}$
This can be written as $x=t^{-2}(1+\log t)$
Differentiating w.r.t $x$ we get,(By applying product rule)
Let $u=t^{-2}$
$\large\frac{du}{dt}=$$-2t^{-3} Let v=(1+\log t) \large\frac{dv}{dt}=\frac{1}{t} Therefore \large\frac{dx}{dt}=$$u.\large\frac{dv}{dt}+v.\large\frac{du}{dt}$
$\qquad\qquad\;\;\;\;=t^{-2}\big(\large\frac{1}{t}\big)$$+(1+\log t).(-2t^{-3}) \qquad\qquad\;\;\;\;=\large\frac{1}{t^3}-\frac{2}{t^3}$$(1+\log t)$
$\qquad\qquad\;\;\;\;=\large\frac{-1}{t^3}$$\big(1+2\log t) Step 2: Consider y=\large\frac{3+2\log t}{t} This can be written as y=(t^{-1})(3+2\log t) Now differentiating w.r.t t by applying product rule, Let u=t^{-1} \large\frac{du}{dt}=-t^{-2} Let v=3+2\log t \large\frac{dv}{dt}$$=2.\large\frac{1}{t}$
$\large\frac{dy}{dt}$$=t^{-1}.3\big(\large\frac{1}{t}\big)$$+(3+2\log t).1$
$\quad\;\;\;=\large\frac{-(3+2\log t)}{t^2}+\frac{2}{t^2}$
$\quad\;\;\;=\large\frac{-1}{t^2}$$(1+2\log t) Step 3: Therefore \large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
$\qquad\qquad=\large\frac{-1/t^2(1+2\log t)}{-1/t^3(1+2\log t)}$
$\qquad\qquad=t$