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3.3: Implicit Differentiation

Difficulty Level: 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*}

The relation, \begin{align*}x^2 + y^2 = 36\end{align*}, in its current form implicitly defines two functions, \begin{align*}f_1(x) = y \end{align*} and \begin{align*} f_2(x) = y\end{align*}. Find these two functions by solving \begin{align*}x^2 + y^2 = 36\end{align*} for \begin{align*}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*} and \begin{align*}f_2(x)\end{align*} explicitly defines the relation \begin{align*}x^2 + y^2 = 36\end{align*}.

Graph \begin{align*}f_1(x)\end{align*} and \begin{align*}f_2(x)\end{align*} on the same set of axis and then draw it in the space to the right. Imagine that you were asked to find the slope of the curve at \begin{align*}x = 2\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*} located on the curve is by taking the derivative of \begin{align*}f_1(x)\end{align*} and \begin{align*}f_2(x)\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*} for \begin{align*}x\end{align*} to determine the slope of the tangents to \begin{align*}x^2 + y^2 = 36\end{align*} at \begin{align*}x = 2\end{align*}. \begin{align*}\frac{dy}{dx}f_1(2)= && \frac{dy}{dx}f_2(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*} using 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*} to find the derivative.

\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*} at \begin{align*}x = 2\end{align*}. First you will need to find the \begin{align*}y-\end{align*}values when \begin{align*}x = 2\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*}. Show that, for all values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, the derivatives of \begin{align*}f_1(x)\end{align*} and \begin{align*}f_2(x)\end{align*} 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*} by Hand

To find the derivative of a relation \begin{align*}F(x, y)\end{align*}, take the derivative of \begin{align*}y\end{align*} with respect to \begin{align*}x\end{align*} of each side of the relation. Looking at the original example, \begin{align*}x^2 + y^2 = 36\end{align*}, we get:

\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*}. Set up the expression up as \begin{align*}\frac{d}{dx}(y(x)^2) \end{align*}. Notice that \begin{align*}y(x)\end{align*} is used rather than just \begin{align*}y\end{align*}. This is very important because it reminds the calculator that \begin{align*}y\end{align*} is a function of \begin{align*}x\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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TI.MAT.ENG.SE.1.Calculus.3.3