TBIL-Calc Differentiating implicitly defined functions (DF7)

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Section 2.7 Differentiating implicitly defined functions (DF7)

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Learning Outcomes

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Compute derivatives of implicitly-defined functions.

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Subsection 2.7.1 Activities

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Observation 2.7.1.

Many of the equations that has been discussed so far fall under the category of an explicit equation. An explicit equation is one in which the relationship between

x

and

y

is given explicitly, such as

y = f (x) .

In this section we will examine when the relationship between

x

and

y

is given implicity. An implicit equation looks like

f (x, y) = g (x, y)

where both sides of the equation may depend on both

x

and

y .

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Observation 2.7.2.

Note that if we are taking the derivative of

f (x)

with respect to

x,

then

\frac{d}{d x} (f (x)) = f^{'} (x) .

However, if we are taking the derivative of

g (y (x))

with respect to

x,

then

\frac{d}{d x} (g (y)) = g^{'} (y) \cdot \frac{d y}{d x} .

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Activity 2.7.3.

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For this activity we want to find the equation of a tangent line for a circle with radius 5 centered at the origin,

x^{2} + y^{2} = 25,

at the point

(- 3, - 4) .

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(a)

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The derivative with respect to

x

for the equation of the circle is given by which expression.

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$2 x + 2 y \frac{d y}{d x} = 25$
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$2 x + y \frac{d y}{d x} = 0$
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$2 x + 2 y \frac{d y}{d x} = 0$
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$2 x + 2 \frac{d y}{d x} = 25$

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(b)

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Solving for

\frac{d y}{d x}

gives?

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$\frac{d y}{d x} = \frac{25 - 2 x}{2 y}$
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$\frac{d y}{d x} = - \frac{2 x}{y}$
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$\frac{d y}{d x} = - \frac{x}{y}$
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$\frac{d y}{d x} = \frac{25 - 2 x}{2}$

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(c)

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Plug the point

(- 3, - 4)

into the expression found above for the derivative to get the slope of the tangent line.

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(d)

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Use the value for the slope of the tangent line to obtain the equation of the tangent line.

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Activity 2.7.4.

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The curve given in Figure 50 is an example of an astroid. The equation of this astroid is

x^{2 / 3} + y^{2 / 3} = 3^{2 / 3} .

What is the derivative with respect

x

for this astroid? (Solve for

\frac{d y}{d x}

Figure 50. Graph of $x^{2 / 3} + y^{2 / 3} = 3^{2 / 3} .$

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$\frac{d y}{d x} = \frac{x^{- 1 / 3}}{y^{- 1 / 3}}$
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$\frac{d y}{d x} = \frac{y^{- 1 / 3}}{x^{- 1 / 3}}$
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$\frac{d y}{d x} = \frac{3^{- 1 / 3} - x^{- 1 / 3}}{y^{- 1 / 3}}$
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$\frac{d y}{d x} = - \frac{x^{- 1 / 3}}{y^{- 1 / 3}}$

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Activity 2.7.5.

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An example of a lemniscate is given in Figure 51. The equation of this lemniscate is

(x^{2} + y^{2})^{2} = x^{2} - y^{2} .

What is the derivative with respect

x

for this lemniscate? (Solve for

\frac{d y}{d x}

Figure 51. Graph of $(x^{2} + y^{2})^{2} = x^{2} - y^{2} .$

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$\frac{d y}{d x} = \frac{x (1 - 2 (x^{2} + y^{2}))}{y + 2 (x^{2} + y^{2})}$
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$\frac{d y}{d x} = \frac{x (1 - 2 (x^{2} + y^{2}))}{y (1 + 2 (x^{2} + y^{2}))}$
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$\frac{d y}{d x} = \frac{y (1 + 2 (x^{2} + y^{2}))}{x (1 - 2 (x^{2} + y^{2}))}$
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$\frac{d y}{d x} = \frac{y + 2 (x^{2} + y^{2})}{x (1 - 2 (x^{2} + y^{2}))}$

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Activity 2.7.6.

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Explain how to use implicit differentiation to find

\frac{d y}{d x}

for each of the following equations.

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(a)

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- 5 x^{5} - 5 \cos (y) = 3 y^{4} + 2

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(b)

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- 5 y e^{x} + 5 \sin (x) = 0

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Activity 2.7.7.

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To take the derivative of some explicit equations you might need to make it an implicit equation. For this activity we will find the derivative of

y = x^{x} .

Make the equation an implicit equation by taking natural logarithm of both sides, this gives

\ln (y) = x \ln (x) .

Knowing this, what is

\frac{d y}{d x} ?

This process to find a derivative is known as logarithmic differentiation.

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$\frac{d y}{d x} = x^{x} (\ln (x) + 1)$
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$\frac{d y}{d x} = \frac{(\ln (x) + 1)}{x^{x}}$
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$\frac{d y}{d x} = x^{x} (\ln (x) + x)$
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$\frac{d y}{d x} = \frac{(\ln (x) + x)}{x^{x}}$

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Activity 2.7.8.

Valerie is building a square chicken coop with side length

x .

Because she needs a separate place for the rooster, she needs to put fence around the square and also along the diagonal line shown. The fence costs $20 per linear meter, and she has a budget of $900.