# Calculus – Tangent Line at a point using Implicit Differentiation

By | October 31, 2016

## Implicit Differentiation

In this tutorial students learn how to find the equation of the tangent line given the point of tangency. The students then verify the solution using the GRAPH, WINDOW, CALC, dy/dx and TRACE features of the TI-84C.

## Implicit Function

$x^3+y^3=9$

## Implicit Differentiation

$y'=\frac {dy}{dx}=\frac {-x^2}{y^2}$

## Substitute in point (2,1) to calculate the slope

$y'=\frac {dy}{dx}=\frac {-(2)^2}{(1)^2}=\frac {-4}{1}=-4$

## Tangent Line

$y=-4x+9$

1) If only the x value is given we should first find the point of tangency by inputting that x value into the function.

2) Calculate the first derivative of the function. This expression represents the slope of the tangent line.

3) Input the x value into the first derivative to get the numerical value of the first derivative (slope).

4a) Use the point slope formula to find the equation of the tangent line.

y-y1=m(x-x1), where the point of tangency is (x1,y1) and the slope is m.

4b) Substitute in the values for x1, y1, and m.

5) Verify your results using the Texas Instrument Graphing Calculator. TI-84, TI-84 Plus, TI-84 C

You have just found the equation of the line tangent to the function at the specified x value.

## Graphing Tool

Click here to view this question using a online interactive graphing calculator.