Question

Find dy/dx if 2y^3t+t^3y=1 and dt/dx=1/cost

Answer 1

Given $2y^{3t}+t^{3y}=1$ and $\frac{dt}{dx}=\frac{1}{cost}$.

Let be $F(y;t)=2y^{3t}+t^{3y}-1$

So $\frac{dy}{dt}=-\frac{\frac{dF}{dt}}{\frac{dF}{dy}}$ . (1)

$\frac{dF}{dt}=2y^{3t}*3lny+3yt^{3y-1}$

$\frac{dF}{dt}=6y^{3t}lny+3yt^{3y-1}$ (2)

$\frac{dF}{dy}=2*3ty^{3t-1}+3t^{3y}lnt$

$\frac{dF}{dy}=6ty^{3t-1}+3t^{3y}lnt$ (3)

Substitute (2) and (3) into (1).

$\frac{dy}{dt}=-\frac{6y^{3t}lny+3yt^{3y-1}}{6ty^{3t-1}+3t^{3y}lnt}$ (4)

Let's find $\frac {dy}{dx}$:

$\frac {dy}{dx}=\frac{dy}{dt}*\frac{dt}{dx}$, (5)

where $\frac{dt}{dx}=\frac{1}{cost}$. (6)

Substitute (4) and (6) into (5):

$\frac {dy}{dx}=-\frac{6y^{3t}lny+3yt^{3y-1}}{6ty^{3t-1}+3t^{3y}lnt}*\frac{1}{cost}$

$\frac {dy}{dx}=-\frac{6y^{3t}lny+3yt^{3y-1}}{cost(6ty^{3t-1}+3t^{3y}lnt)}$ (7)

Answer: $\frac {dy}{dx}=-\frac{6y^{3t}lny+3yt^{3y-1}}{cost(6ty^{3t-1}+3t^{3y}lnt)}$.