# 3.3: Implicit Differentiation

**At Grade**Created by: CK-12

*This activity is intended to supplement Calculus, Chapter 2, Lesson 6.*

##
Problem 1 – Finding the Derivative of x2+y2=36

The relation, *implicitly* defines two functions,

Substitute the above functions in the original relation and then simplify.

This confirms that *explicitly* defines the relation

Graph

- Why might this question be potentially difficult to answer?
- What strategies or methods could you use to answer this question?

One way to find the slope of a tangent drawn to the circle at any point

Check that your derivatives are correct by using the **Derivative** command (press **F3:Calc > 1:d( diffferentiate**) on the *Calculator* screen.

Substitute

Another way to find the slope of a tangent is by finding the derivative of *implicit differentiation*. On the *Calculator* screen press **F3:Calc > D:impDif(** to access the **impDif** command. Enter **impDif**

Use this result to find the slope of the tangents to

- Is your answer consistent with what was found earlier?
- Rewrite the implicit differentiation derivative in terms of
x . Show that, for all values ofx andy , the derivatives off1(x) andf2(x) that you found earlier are equal to the result found using the**impDif**command.

##
Problem 2 – Finding the Derivative of x2+y2=36 by Hand

To find the derivative of a relation

Evaluate the following and by hand.

Use the **Derivative** command to find

You have now evaluated \begin{align*}\frac{d}{dx}(x^2),\frac{d}{dx}(y^2)\end{align*}, and \begin{align*}\frac{d}{dx}(36)\end{align*}. Replace these expressions in the equation \begin{align*}\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(36)\end{align*} and solve for \begin{align*}\frac{dy}{dx}\end{align*}.

Compare your result to the one obtained using the **impDif** command.

## Problem 3 – Finding the Derivative of \begin{align*}y^2 + xy = 2\end{align*}

The relation, \begin{align*}y^2 + xy = 2\end{align*}, can also be solved as two functions, \begin{align*}f_1(x)\end{align*} and \begin{align*}f_2(x)\end{align*}, which *explicitly* define it.

- What strategy can be used to solve \begin{align*}y^2 + xy = 2\end{align*} for \begin{align*}y\end{align*}?

Solve \begin{align*}y^2 + xy = 2\end{align*} for \begin{align*}y\end{align*} and use the **Solve** command (press **F2:Algebra > 1:solve(**) to check your answer.

The derivative of \begin{align*}y^2 + xy = 2\end{align*} can then be found by taking the derivatives of \begin{align*}f_1(x)\end{align*} and \begin{align*}f_2(x)\end{align*}. However, the derivative can be found more easily using implicit differentiation.

Use implicit differentiation to find the derivative of \begin{align*}y^2 + xy = 2\end{align*}. Check your result by using the **impDif** command. (*Hint:* The product rule must be used to find the derivative of \begin{align*}xy\end{align*}.)

\begin{align*} \frac{dy}{dx}= \end{align*}

Use the derivative you found for \begin{align*}y^2 + xy = 2\end{align*} to calculate the slope at \begin{align*}x = -6\end{align*}. First you will need to find the \begin{align*}y-\end{align*}values when \begin{align*}x = -6\end{align*}.

\begin{align*} \frac{dy}{dx}(-6,y)= && \frac{dy}{dx}(-6,y)= \end{align*}

Verify your result graphically. Graph the two functions, \begin{align*}f_1(x)\end{align*} and \begin{align*}f_2(x)\end{align*}. Then use the slopes and points to graph each tangent line.

##
Extension – Finding the Derivative of **\begin{align*}x^3 + y^3 = 6xy\end{align*}**

The relation \begin{align*}x^3 + y^3 = 6xy\end{align*} cannot be solved explicitly for \begin{align*}y\end{align*}. In this case implicit differentiation must be used.

- Find the derivative of \begin{align*}x^3 + y^3 = 6xy\end{align*} and use the
**impDif**command to verify your result.

\begin{align*} \frac{dy}{dx}= \end{align*}

Use this result to find the slope of the tangents to \begin{align*}x^3 + y^3 = 6xy\end{align*} at \begin{align*}x = 1\end{align*}. (*Hint*: Use the **solve** command to find the \begin{align*}y\end{align*} values that correspond to \begin{align*}x = 1\end{align*}.)

### My Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes |
---|---|---|

Show More |