Solution assignment 06 Implicit differentiation

Calculate the tangent line at the point $(1,3)$ of the graph of the following expression:

$2x+3y^2=29$

Implicit differentiation in combination with the chain rule yields:

$2+6yy'=0$ , so:

$y'=-\displaystyle\frac{1}{3y}$

The derivative in $(1,3)$ is:

$y'=-\displaystyle\frac{1}{9}$

The tangent line through the point $(1,3)$ is:

$y-3=-\displaystyle\frac{1}{9}(x-1)$

or:

$y=-\displaystyle\frac{1}{9}x+3\displaystyle\frac{1}{9}$