How to Find the Equation of a Tangent Line

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Last updated on September 25th, 2020

The tangent line is a straight line with that slope, passing through that exact point on the graph. To find the equation for the tangent, you’ll need to know how to take the derivative of the original equation. Here is how to find the equation of a tangent line.

• To find the equation of a line you need a point and a slope.
• The slope of the tangent line is the value of the derivative at the point of tangency.

Steps to find the equation of a tangent line

1. Find the first derivative of f(x).

Suppose $\dpi{120}&space;\large&space;f(x)&space;=&space;x^{3}$. Find the equation of the tangent line at the point where x = 2.

Recall the power rule when taking derivatives: $\dpi{120}&space;\large&space;\frac{d}{dx}x^{n}&space;=&space;nx^{n-1}$ .

The function’s first derivative = $\dpi{120}&space;\large&space;{f}'(x)&space;=&space;3x^{2}$

2. The plug x value of the indicated point into f ‘(x) to find the slope at x.

The slope of the tangent line is m = 12

3. Plug x value into f(x) to find the y coordinate of the tangent point.

$\dpi{120}&space;\large&space;f(2)&space;=&space;2^{3}&space;=&space;8$  The point is (2, 8).

4. Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line.
5. Graph your results to see if they are reasonable. Here is the graph of the function and the tangent line we just found.

How to Find the equation of tangent lines to Implicit Curves

1. Find the y-value of the point of tangency.

Suppose $\dpi{120}&space;\large&space;x^{2}+y^{2}&space;=&space;16$Find the equation of the tangent line at x = 2x=2 for y > 0y>0.

$\dpi{120}&space;\large&space;\\x^{2}&space;+&space;y^{2}&space;=&space;16&space;\\2^{2}&space;+&space;y^{2}&space;=&space;16&space;\\&space;4\&space;+&space;y^{2}&space;=&space;16&space;\\y^{2}&space;=&space;16&space;-&space;4&space;\\y^{2}&space;=&space;12&space;\\y\&space;=&space;\pm&space;\sqrt{12}&space;\\y\&space;=&space;\pm&space;\sqrt{4.3}&space;\\y\&space;=&space;\pm&space;2\sqrt{3}$

$\dpi{120}&space;\large&space;y&space;=&space;2\sqrt{3}$ since y > 0. The point of tangency is $\dpi{120}&space;\large&space;(2,&space;\sqrt{3})$

2. Find the equation for $\dpi{120}&space;\large&space;\frac{dx}{dy}$.

The equation is implicitly defined, we use implicit differentiation.

$\dpi{120}&space;\large&space;\\2x&space;+&space;2y&space;\frac{dx}{dy}&space;=&space;0&space;\\2y&space;\frac{dx}{dy}&space;=&space;-2x&space;\\\frac{dx}{dy}&space;=&space;-\frac{2x}{2y}&space;\\\frac{dx}{dy}&space;=&space;-\frac{x}{y}$

3. Find the slope of the tangent line at the point of tangency.

$\frac{\mathrm{d}&space;x}{\mathrm{d}&space;y}|(2,{\color{Red}&space;2\sqrt{3}})&space;=&space;-&space;\frac{2}{{\color{Red}&space;2\sqrt{3}}}&space;=&space;-\frac{1}{\sqrt{3}}&space;=&space;-\frac{1}{\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}&space;=&space;-\frac{\sqrt{3}}{3}$

4. Find the equation of the tangent line. Point $\dpi{120}&space;\large&space;(2,&space;\sqrt{3})$ and
slop m = $-\frac{\sqrt{3}}{3}$

$\\y&space;-&space;y_{1}\&space;\&space;\&space;=&space;x&space;-&space;x_{1}&space;\\y&space;-&space;2\sqrt{3}&space;=&space;-\frac{\sqrt{3}}{3}(x&space;-&space;2)$

The equation of the tangent line is $y&space;-&space;2\sqrt{3}&space;=&space;-\frac{\sqrt{3}}{3}(x&space;-&space;2)$

For reference, the graph of the curve and the tangent line we found is shown below.