91Ó°ÊÓ

They provide an equation or curve that is $x{y}^3-y{x}^3=6$

To find the slopes of tangents to curves that aren't functions, we can utilize implicit differentiation (they fail the vertical line test). Parts of yare assumed to be functions that satisfy the supplied equation but yis not a function of x.

Consider the equation of the curve below:

$x{y}^3-y{x}^3=6$

Differentiate the provided above equation of a curve,

$$\begin{gathered} x.3y^3\frac{dy}{dx}+y^3-\left(y.3x^2+x^3.\frac{dy}{dx}\right)=0 \\ 3x{y}^3\frac{dy}{dx}+y^3-3x^{2}y-x^3.\frac{dy}{dx}=0 \\ {y}^3-3x^{2}y+\left(3x{y}^2-x^3\right)\frac{dy}{dx}=0 \end{gathered}$$

Hence,

$\frac{dy}{dx}=\frac{3x^{2}y-y^3}{3x{y}^2-x^3}$

Again, differentiate the provided above equation of a curve,

role="math" localid="1658814580023" $$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{\begin{array}{c}\left(3x^{2}y-y^3\right)\left[3\left(x^2{\displaystyle \frac{dy}{dx}}+2xy\right)-3y^2{\displaystyle \frac{dy}{dx}}\right]\\ -\left(3x^2-y-y^3\right)\left[3\left(2xy{\displaystyle \frac{dy}{dx}}+y^2\right)-3x^2\right]\end{array}}{{\left(3x{y}^2-x^3\right)}^2} \\ =\frac{\begin{array}{c}\left(3x{y}^2-x^3\right)\left[3x^2\left({\displaystyle \frac{3x^{2}y-y^3}{x\left(3y^2-x^2\right)}}\right)+6xy-3y^2\left({\displaystyle \frac{3x^{2}y-y^3}{3y^2-x^2}}\right)\right]\\ -\left(3x^2-y-y^3\right)\left[6xy\left({\displaystyle \frac{3x^{2}y-y^3}{x\left(3y^2-x^2\right)}}\right)+3y^2-3x^2\right]\end{array}}{{\left(3x{y}^2-x^3\right)}^2} \end{gathered}$$

Put the (1,2) for (x,y)

$$\begin{gathered} {\left(\frac{d^{2}y}{d{x}^2}\right)}_{\left(1,2\right)}=\frac{\begin{array}{c}3\left(1\right){\left(2\right)}^2-{\left(1\right)}^{3}3{\left(1\right)}^2\left[\begin{array}{c}\left(\frac{\left(3{\left(1\right)}^2\left(2\right)-{\left(2\right)}^3\right)}{\left(1\right)\left(3{\left(2\right)}^2-{\left(1\right)}^2\right)}\right)6\left(1\right)\left(2\right)\\ -3{\left(2\right)}^2\left(\frac{3{\left(1\right)}^2\left(2\right)-{\left(2\right)}^3}{3{\left(2\right)}^2-{\left(1\right)}^2}\right)\end{array}\right]\\ \left(-3{\left(1\right)}^2\left(2\right)-{\left(2\right)}^3\right)\left[\begin{array}{c}6\left(1\right)\left(2\right)\left({\displaystyle \frac{3{\left(1\right)}^2\left(2\right)-{\left(2\right)}^3}{\left(1\right)\left(3{\left(2\right)}^2-{\left(1\right)}^2\right)}}\right)\\ +3{\left(2\right)}^2-3{\left(1\right)}^2\end{array}\right]\\ \end{array}}{{\left(3\left(1\right){\left(2\right)}^2-{\left(1\right)}^3\right)}^2} \\ =\frac{1800}{1331} \end{gathered}$$