# 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 \begin{align*}x^2 + y^2 = 36\end{align*}x2+y2=36

The relation, \begin{align*}x^2 + y^2 = 36\end{align*}*implicitly* defines two functions, \begin{align*}f_1(x) = y \end{align*}

\begin{align*}f_1(x) = && f_2(x) =\end{align*}

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

\begin{align*}x^2 + (f_1(x))^2 = 36 && x^2 + (f_2(x))^2 = 36\end{align*}

This confirms that \begin{align*}f_1(x)\end{align*}*explicitly* defines the relation \begin{align*}x^2 + y^2 = 36\end{align*}

Graph \begin{align*}f_1(x)\end{align*}

- 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 \begin{align*}(x,y)\end{align*}

\begin{align*}\frac{dy}{dx}f_1(x)= && \frac{dy}{dx}f_2(x)=\end{align*}

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

Substitute \begin{align*}2\end{align*}

Another way to find the slope of a tangent is by finding the derivative of \begin{align*}x^2 + y^2 = 36\end{align*}*implicit differentiation*. On the *Calculator* screen press **F3:Calc > D:impDif(** to access the **impDif** command. Enter **impDif** \begin{align*}(x^2 + y^2 = 36,x, y)\end{align*}

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

Use this result to find the slope of the tangents to \begin{align*}x^2 + y^2 = 36\end{align*}

\begin{align*}\frac{dy}{dx}(2,y)= && \frac{dy}{dx}(2,y)=\end{align*}

- Is your answer consistent with what was found earlier?
- Rewrite the implicit differentiation derivative in terms of \begin{align*}x\end{align*}
x . Show that, for all values of \begin{align*}x\end{align*}x and \begin{align*}y\end{align*}y , the derivatives of \begin{align*}f_1(x)\end{align*}f1(x) and \begin{align*}f_2(x)\end{align*}f2(x) that you found earlier are equal to the result found using the**impDif**command.

##
Problem 2 – Finding the Derivative of \begin{align*}x^2 + y^2 = 36\end{align*}x2+y2=36 by Hand

To find the derivative of a relation \begin{align*}F(x, y)\end{align*}

\begin{align*}\frac{d}{dx}(x^2 + y^2) &= \frac{d}{dx}(36)\\
\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) &= \frac{d}{dx}(36) \end{align*}

Evaluate the following and by hand.

\begin{align*}\frac{d}{dx}(x^2) = && \frac{d}{dx}(36)= \end{align*}

Use the **Derivative** command to find \begin{align*}\frac{d}{dx}(y^2)\end{align*}

\begin{align*}\frac{d}{dx}(y^2) = \end{align*}

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*}.)

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