Question

suppose that y^3-2xy=4x^3a) use implicit differentiation to find dy/dxb) find an equation for the line tangent to the curve at (1,2)

Answer 1

`\begin{gathered} y^3-2xy=4x^3 \\ \end{gathered}`

As a first step to implicit differentiation, we are going to differentite both sides of the equation:

`\begin{gathered} (d)/(dx)(y^3-2xy)=(d)/(dx)(4x^3) \\ (d)/(dx)(y^3)-(d)/(dx)(2xy)=12x^2 \\ \text{Take the derivates:} \\ 3y^2(dy)/(dx)-2y(dy)/(dx)=12x^2 \end{gathered}`

Now, we are going to keep the terms with dy/dx on the left and move the remaining terms to the right:

`\begin{gathered} 3y^2(dy)/(dx)-2y(dy)/(dx)=12x^2 \\ (dy)/(dx)(3y^2-2y)=12x^2 \\ (dy)/(dx)=(12x^2)/(3y^2-2y) \end{gathered}`

The equation of the tangent line can be find by substituing (1, 2) into the expression dy/dx, to find the slope:

`\begin{gathered} \text{slope}=(12(1)^2)/(3(2)^2-2(2))=(12)/(8)=(3)/(2) \\ \end{gathered}`

Using the point given and the slope we just calculate, we can find the equation in the point-slope form and then solve for y to give the general equation:

`\begin{gathered} y-2=(3)/(2)(x-1) \\ y=(3)/(2)x-(3)/(2)+2 \\ y=(3)/(2)x+(1)/(2) \end{gathered}`