# Assume x and y are functions of t. Evaluate \frac{dy}{dt} \ for \ 3xy - 2xy = 4 y^{3} = -152,...

## Question:

Assume x and y are functions of t. Evaluate {eq}\frac{dy}{dt} \ for \ 3xy - 2xy = 4 y^{3} = -152 {/eq}, with the conditions {eq}\frac{dx}{dy} = -12, x-4, y=-3. {/eq}

\frac{dy}{dt} =

## Differentiation of the Function:

Differentiation is the process of finding the derivative of a function.The given function is in the product form of two function {eq}\displaystyle (u\cdot v). {/eq} To find the derivative of the given we use the product Rule of the differentiation of the function.

{eq}\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(u\cdot v)=u\cdot \frac{\mathrm{d} v}{\mathrm{d} x}+v\cdot \frac{\mathrm{d} u}{\mathrm{d} x} {/eq}

Consider the function

{eq}\displaystyle 3xy - 2xy-4 y^{3} = -152,\: \: \frac{\mathrm{d} x}{\mathrm{d} t}=-12,\: \: x=4,\: \: y=-3\\ \displaystyle xy-4 y^{3} = -152 {/eq}

Differentiate with respect to t, we get

{eq}\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(xy-4 y^{3} )= 0\\ \displaystyle (x \frac{dy}{dt}+y \frac{dx}{dt}) - 12 y^{2} \frac{\mathrm{d} y}{\mathrm{d} t}= 0\\ \displaystyle (x-12y^2) \frac{dy}{dt}+y \frac{dx}{dt}= 0\\ \displaystyle (x-12y^2) \frac{dy}{dt}=-y \frac{dx}{dt}\\ \displaystyle \frac{dy}{dt}=-\frac{y}{(x-12y^2)}\cdot \frac{dx}{dt}\\ \displaystyle \frac{dy}{dt}=\frac{y}{(12y^2-x)}\cdot \frac{dx}{dt}\\ \displaystyle \left ( \frac{dy}{dt} \right )\Biggr |_{ \frac{\mathrm{d} x}{\mathrm{d} t}=-12,\: \: x=4,\: \: y=-3} =\frac{-3}{(12(-3)^2-4)}\cdot (-12)=\frac{9}{26}\\ {/eq} 