3.8 Implicit differentiation  (Page 2/4)

Follow the steps in the problem-solving strategy.

In [link] , we showed that $\frac{dy}{dx}=-\frac{x}{y}.$ We can take the derivative of both sides of this equation to find $\frac{d^{2}y}{d{x}^2}.$

At this point we have found an expression for $\frac{d^{2}y}{d{x}^2}.$ If we choose, we can simplify the expression further by recalling that $x^2+y^2=25$ and making this substitution in the numerator to obtain $\frac{d^{2}y}{d{x}^2}=-\frac{25}{y^3}.$

Although we could find this equation without using implicit differentiation, using that method makes it much easier. In [link] , we found $\frac{dy}{dx}=-\frac{x}{y}.$

The slope of the tangent line is found by substituting $\left(3,\mathrm{-4}\right)$ into this expression. Consequently, the slope of the tangent line is $\frac{dy}{dx}|\begin{array}{l}\\ {}_{\left(3,\mathrm{-4}\right)}\end{array}=-\frac{3}{\mathrm{-4}}=\frac{3}{4}.$

Using the point $\left(3,\mathrm{-4}\right)$ and the slope $\frac{3}{4}$ in the point-slope equation of the line, we obtain the equation $y=\frac{3}{4}x-\frac{25}{4}$ ( [link] ).

Begin by finding $\frac{dy}{dx}.$

Next, substitute $\left(\frac{3}{2},\frac{3}{2}\right)$ into $\frac{dy}{dx}=\frac{3y-3x^2}{3y^2-3x}$ to find the slope of the tangent line:

Finally, substitute into the point-slope equation of the line to obtain