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

Sep 20, 2014

Let us this example:

Find the equation of the tangent line to the circle ${x}^{2} + {y}^{2} = {5}^{2}$ at the point $\left(3 , 4\right)$. 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,

$\frac{d}{\mathrm{dx}} \left({x}^{2} + {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left({5}^{2}\right) R i g h t a r r o w 2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

by dividing by $2 y$,

$\frac{x}{y} + \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

by subtracting $\frac{x}{y}$,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

So, we can find $m$ by evaluating $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\left(3 , 4\right)$.

$m = \frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\left(3 , 4\right)} = - \frac{3}{4}$

By Point-Slope Form: $y - {y}_{1} = m \left(x - {x}_{1}\right)$,

Tangent Line: $y - 4 = - \frac{3}{4} \left(x - 3\right)$