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:

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

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

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

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

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

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:

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