Question

How do you find the Tangent line to a curve by implicit differentiation?

Answer 1

Let us this example:

Find the equation of the tangent line to the circle `x^2+y^2=5^2` at the point `(3,4)`. In order to identify a line, we need two pieces of information:
`{("Point: " (x_1,y_1)=(3,4)), ("Slope: " m=?):}`

Since the point is already provided, all you need is the slope `m`.
Let us find `m` by implicit differentiation.

By implicitly differentiating,

`d/{dx}(x^2+y^2)=d/{dx}(5^2)Rightarrow 2x+2y{dy}/{dx}=0`

by dividing by `2y`,

`{x}/{y}+{dy}/{dx}=0`

by subtracting `x/y`,

`{dy}/{dx}=-x/y`

So, we can find `m` by evaluating `{dy}/{dx}` at `(3,4)`.

`m={dy}/{dx}|_{(3,4)}=-3/4`

By Point-Slope Form: `y-y_1=m(x-x_1)`,

Tangent Line: `y-4=-3/4(x-3)`