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3.8 Implicit Differentiation

A function is in explicit form if \( y \) is on its own side and by itself.

e.g. \( y = x^2 - 5x + 10 \)

\( y \) is on its own side, not raised to a power other than 1 and not part of another function (e.g. \( \sin(y) \))

A function is in implicit form if it is not in explicit form.

e.g. \( x^2 + y^2 = 9 \)

\( xy = \sin(xy) \)

Sometimes we can turn implicit into explicit:

\( x^2 + y^2 = 9 \)

\( y = \sqrt{9 - x^2} \)

but it's not always possible:

\( xy = \sin(xy) \)

\( y \) cannot be isolated

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Finding \( \frac{dy}{dx} \) or \( y' \) when \( y \) is explicit is straightforward.

But if \( y \) is implicit (e.g. \( xy = \sin(xy) \)), we need implicit differentiation.

Implicit differentiation is built upon the Chain Rule.

If we know \( y = 3x^2 + 5 \)

then \( \frac{d}{dx}(y)^3 = \frac{d}{dx}(3x^2 + 5)^3 = 3 \cdot (3x^2 + 5)^2 \cdot (6x) \)

Note: \( u = 3x^2 + 5 \), so the derivative is \( 3u^2 \cdot \frac{du}{dx} \)

The same applies if we don't know what \( y \) looks like but we know \( y \) is a function of \( x \).

\( \frac{d}{dx}(y)^3 = 3y^2 \cdot \frac{dy}{dx} \)

\( \frac{dy}{dx} \) comes from the chain rule.

\( (y)^3 \) is some function of \( y \) and \( x \).

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Implicit Differentiation Example

Example

Given the equation: \[ x^3 + y^3 = 8 \] Find \( \frac{dy}{dx} \)

Note: \( y \) is some implicit function of \( x \)

Take the derivative with respect to \( x \) on both sides:

\[ \frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(8) \]
\[ \frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(8) \]

Since \( y \) is a function of \( x \), the chain rule applies.

\[ 3x^2 + 3y^2 \frac{dy}{dx} = 0 \]

Now solve for \( \frac{dy}{dx} \):

\[ 3y^2 \frac{dy}{dx} = -3x^2 \]

\[ \frac{dy}{dx} = -\frac{x^2}{y^2} \]

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We can actually rewrite \( x^3 + y^3 = 8 \) in explicit form:

\[ y^3 = 8 - x^3 \] \[ y = (8 - x^3)^{1/3} \]

Its derivative should match the other one:

\[ y' = \frac{dy}{dx} = \frac{1}{3}(8 - x^3)^{-2/3} \cdot \frac{d}{dx}(8 - x^3) \] \[ = \frac{1}{3}(8 - x^3)^{-2/3}(-3x^2) = \frac{-x^2}{(8 - x^3)^{2/3}} \]

This is also:

\[ \frac{-x^2}{[(8 - x^3)^{1/3}]^2} = \frac{-x^2}{y^2} \]

Matches the other form.

Note: In the denominator, \( (8 - x^3)^{1/3} \) is substituted back as \( y \).

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Example

\[ \frac{1}{x} + \frac{1}{y} = 2 \quad \text{find } \frac{dy}{dx} \]

y is some function of x

rewrite:

\[ x^{-1} + y^{-1} = 2 \]

take deriv. of both sides with respect to x

\[ \frac{d}{dx}(x^{-1}) + \frac{d}{dx}(y^{-1}) = \frac{d}{dx}(2) \]

y contains x so Chain Rule applies

\[ -x^{-2} + (-y^{-2} \frac{dy}{dx}) = 0 \]

solve for \( \frac{dy}{dx} \)

\[ -y^{-2} \frac{dy}{dx} = x^{-2} \]
\[ \frac{dy}{dx} = \frac{x^{-2}}{-y^{-2}} = \boxed{-\frac{y^2}{x^2}} \]
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Example

\[ xy = \sin(xy) \]

contains x

differentiate both sides with respect to x

\[ \frac{d}{dx}(xy) = \frac{d}{dx}[\sin(xy)] \]

product of two functions of x

product rule is required

\[ (x) \cdot \frac{d}{dx}(y) + (y) \cdot \frac{d}{dx}(x) = \frac{d}{dx}[\sin(xy)] \]
\[ x \cdot \frac{dy}{dx} + y \cdot 1 = \cos(xy) \cdot \frac{d}{dx}(xy) \]

product rule

\[ x \frac{dy}{dx} + y = \cos(xy) \cdot (x \cdot \frac{dy}{dx} + y) \]

solve for \( \frac{dy}{dx} \)

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\[ x \frac{dy}{dx} + y = x \cos(xy) \frac{dy}{dx} + y \cos(xy) \]
\[ x \frac{dy}{dx} - x \cos(xy) \frac{dy}{dx} = y \cos(xy) - y \]
\[ \frac{dy}{dx} [x - x \cos(xy)] = y \cos(xy) - y \]
\[ \frac{dy}{dx} = \frac{y \cos(xy) - y}{x - x \cos(xy)} \]
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example

\[ e^{2y} + x = 8y \quad \text{find } \frac{d^2y}{dx^2} \]

Contain x

derivative with respect to x on both sides

\[ \frac{d}{dx}(e^{2y}) + \frac{d}{dx}(x) = \frac{d}{dx}(8y) \]
\[ \frac{d}{dx}(e^u) = e^u \cdot \frac{du}{dx} \]
\[ e^{2y} \cdot \frac{d}{dx}(2y) + 1 = 8 \frac{dy}{dx} \]
\[ e^{2y} \cdot 2 \frac{dy}{dx} + 1 = 8 \frac{dy}{dx} \]

solve for \( \frac{dy}{dx} \)

\[ 2e^{2y} \frac{dy}{dx} - 8 \frac{dy}{dx} = -1 \]
\[ \frac{dy}{dx} (2e^{2y} - 8) = -1 \]
\[ \frac{dy}{dx} = \frac{-1}{2e^{2y} - 8} \]

now differentiate again to find \( \frac{d^2y}{dx^2} \)

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Second Derivative Calculation

\[\frac{dy}{dx} = -(2e^{2y} - 8)^{-1}\]

The expression is in the form of \(u^{-1}\), where the derivative is \(-u^{-2} \cdot \frac{du}{dx}\).

Step-by-Step Differentiation

\[\begin{aligned} \frac{d^2y}{dx^2} &= -(-1)(2e^{2y} - 8)^{-2} \cdot \frac{d}{dx}(2e^{2y} - 8) \\ &= (2e^{2y} - 8)^{-2} \cdot 2 \cdot e^{2y} \cdot \frac{d}{dx}(2y) \\ &= (2e^{2y} - 8)^{-2} \cdot 2 \cdot e^{2y} \cdot 2 \frac{dy}{dx} \\ &= 4e^{2y}(2e^{2y} - 8)^{-2} \cdot \left[ \frac{dy}{dx} \right] \end{aligned}\]

Note: We know what the term \(\frac{dy}{dx}\) looks like from the initial equation.

\[\begin{aligned} &= 4e^{2y}(2e^{2y} - 8)^{-2} \cdot -(2e^{2y} - 8)^{-1} \\ &= -4e^{2y} \cdot (2e^{2y} - 8)^{-3} \end{aligned}\]

Final Result

\[\frac{d^2y}{dx^2} = \frac{-4e^{2y}}{(2e^{2y} - 8)^3}\]