A curve C is defined by the parametric equations x=(4-e^(2-6t))/4 , y=e^(3t)/(3t), t doesnt = 0. Find the exact value of dy/dx at the point on C where t=2/3 .

To solve this we must use the chain rule which is dy/dt * dt/dx. Firstly, we differentiate dy/dt. For this we must use the quotient rule, this gives us dy/dt=(9te^3t - 3e^3t)/9t^2. Now for dx/dt, by substituting u=2-6t and using the chain rule du/dt * dx/du we get du/dt=-6 and dx/du=-(e^u)/4, we times these together and substitute u=2-6t back in to get dx/dt=(3e^(2-6t))/2. Now we must times dy/dt * 1/(dx/dt) to give us dy/dx=(18te^(3t)-6e^(3t))/(27t^(2)e^(2-6t)). We now must substitute in t=2/3 to give us the exact point on the curve that we require. By doing this and then simplifying our answer we get the value of the gradient dy/dx=(e^4)/2.