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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 $x^2 + y^2 = 36$

The relation, $x^2 + y^2 = 36$, in its current form implicitly defines two functions, $f_1(x) = y$ and $f_2(x) = y$. Find these two functions by solving $x^2 + y^2 = 36$ for $y$.

$f_1(x) = && f_2(x) =$

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

$x^2 + (f_1(x))^2 = 36 && x^2 + (f_2(x))^2 = 36$

This confirms that $f_1(x)$ and $f_2(x)$ explicitly defines the relation $x^2 + y^2 = 36$.

Graph $f_1(x)$ and $f_2(x)$ 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 $x = 2$.

• 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 $(x,y)$ located on the curve is by taking the derivative of $f_1(x)$ and $f_2(x)$.

$\frac{dy}{dx}f_1(x)= && \frac{dy}{dx}f_2(x)=$

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

Substitute $2$ for $x$ to determine the slope of the tangents to $x^2 + y^2 = 36$ at $x = 2$. $\frac{dy}{dx}f_1(2)= && \frac{dy}{dx}f_2(2)=$

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

$\frac{dy}{dx}=$

Use this result to find the slope of the tangents to $x^2 + y^2 = 36$ at $x = 2$. First you will need to find the $y-$values when $x = 2$.

$\frac{dy}{dx}(2,y)= && \frac{dy}{dx}(2,y)=$

• Rewrite the implicit differentiation derivative in terms of $x$. Show that, for all values of $x$ and $y$, the derivatives of $f_1(x)$ and $f_2(x)$ that you found earlier are equal to the result found using the impDif command.

## Problem 2 – Finding the Derivative of $x^2 + y^2 = 36$ by Hand

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

$\frac{d}{dx}(x^2 + y^2) &= \frac{d}{dx}(36)\\\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) &= \frac{d}{dx}(36)$

Evaluate the following and by hand.

$\frac{d}{dx}(x^2) = && \frac{d}{dx}(36)=$

Use the Derivative command to find $\frac{d}{dx}(y^2)$. Set up the expression up as $\frac{d}{dx}(y(x)^2)$. Notice that $y(x)$ is used rather than just $y$. This is very important because it reminds the calculator that $y$ is a function of $x$.

$\frac{d}{dx}(y^2) =$

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

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

## Problem 3 – Finding the Derivative of $y^2 + xy = 2$

The relation, $y^2 + xy = 2$, can also be solved as two functions, $f_1(x)$ and $f_2(x)$, which explicitly define it.

• What strategy can be used to solve $y^2 + xy = 2$ for $y$?

Solve $y^2 + xy = 2$ for $y$ and use the Solve command (press F2:Algebra > 1:solve() to check your answer.

The derivative of $y^2 + xy = 2$ can then be found by taking the derivatives of $f_1(x)$ and $f_2(x)$. However, the derivative can be found more easily using implicit differentiation.

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

$\frac{dy}{dx}=$

Use the derivative you found for $y^2 + xy = 2$ to calculate the slope at $x = -6$. First you will need to find the $y-$values when $x = -6$.

$\frac{dy}{dx}(-6,y)= && \frac{dy}{dx}(-6,y)=$

Verify your result graphically. Graph the two functions, $f_1(x)$ and $f_2(x)$. Then use the slopes and points to graph each tangent line.

## Extension – Finding the Derivative of $x^3 + y^3 = 6xy$

The relation $x^3 + y^3 = 6xy$ cannot be solved explicitly for $y$. In this case implicit differentiation must be used.

• Find the derivative of $x^3 + y^3 = 6xy$ and use the impDif command to verify your result.

$\frac{dy}{dx}=$

Use this result to find the slope of the tangents to $x^3 + y^3 = 6xy$ at $x = 1$. (Hint: Use the solve command to find the $y$ values that correspond to $x = 1$.)

Feb 23, 2012

Nov 04, 2014