Lesson 16 (3.8)

Lesson 16 (3.8)

Today: Implicit Differentiation

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Warmup Example:

\[ y = (\underbrace{\sin x + x^3 - 23}_{u})^3 \quad \text{Find } y' \]

\[ \begin{aligned} y' = \frac{dy}{dx} = \frac{du^3}{dx} &= \frac{du^3}{du} \frac{du}{dx} \\[1em] &= 3u^2 \cdot \frac{du}{dx} \\[1em] &= 3\left[\sin x + x^3 - 23\right]^2 \cdot \frac{d}{dx}\left[\sin x + x^3 - 23\right] \\[1em] &= 3\left[\sin x + x^3 - 23\right]^2 \left[\cos x + 3x^2\right] \end{aligned} \]

in fact, \( f(x) \) is any function

\[ y = (f(x))^3 = 3(f(x))^2 \cdot f'(x) \]

Explicit function v/s Implicit function

\( x = \text{independent} \)
\( y = \text{dependent on } x \)

explicit: \( y = f(x) \) “Solve for \( y \)”
or write dependent \( y \) on one side and all \( x \) on another

eg:
\( y = x^3 \)
\( y = \sin x + \tan x - x^3 \)

implicit: \( x \ \& \ y \) are together in an equation and \( y \) is difficult to solve for.

eg:
\( x^2 + y^3 = 5 \rightsquigarrow \text{Not too difficult to solve for } y. \)
\( \sin(xy) + y^3 - 2x^3y = 7y^2 \rightsquigarrow y \text{ is a function of } x \)
\(\text{how to find } y'? \)

Example: \(x^2 + y^3 = 5\), find \(\frac{dy}{dx}\)

We could actually write \(y\) explicitly

\[ y = (5-x^2)^{1/3} \] \[ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}\left[5-x^2\right]^{1/3} \\ &= \frac{1}{3}\left[5-x^2\right]^{1/3-1} \frac{d}{dx}\left[5-x^2\right] \\ &= \frac{1}{3}\left[5-x^2\right]^{-2/3}[-2x] = \frac{-2x}{3(5-x^2)^{2/3}} \end{aligned} \]

\(y = (5-x^2)^{1/3}\)

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

\[ x^2 + y^3 = 5 \rightarrow x^2 + (y(x))^3 = 5 \]

Instead of solving for \( y(x) \) & taking derivative take \( \frac{d}{dx} \) on both sides of equation

\[ \frac{d}{dx} \left( x^2 + (y(x))^3 \right) = \frac{d}{dx}(5) \] \[ 2x + \frac{d}{dx} [y(x)]^3 = 0 \]

use Chain Rule   \( \frac{d}{dx} [y(x)]^3 = 3[y(x)]^2 \cdot y'(x) \)

\[ 2x + 3(y(x))^2 \cdot y'(x) = 0 \]

Solve for \( y'(x) \)

\[ y'(x) = \frac{-2x}{3y^2} \]

Implicit Differentiation

\(y\) is a function of \(x\), implicitly given to find \(y'\)

1. Take \(\frac{d}{dx}\) on both sides of the equation
2. Apply Chain Rule where ever you see \(y\).
   get an equation relating \(x, y, y'\).
3. Solve for \(y'\)

eg: \( x^2 + xy - y^3 = 7 \quad \text{Find } y' \)

\[ x^2 + x(y(x)) - (y(x))^3 = 7 \]

1. \( \frac{d}{dx} \) on both sides of equation

   \[ \frac{d}{dx}\left[x^2 + xy - y^3\right] = \frac{d}{dx} 7 \] \[ \frac{d}{dx}(x^2) + \frac{d}{dx}\left[x(y(x))\right] - \frac{d}{dx}[y(x)]^3 = 0 \]

2. Apply chain Rule wherever there is \( y(x) \)

   \[ 2x + \left( x \cdot \frac{dy}{dx} + \frac{dx}{dx} y \right) - \left( 3[y(x)]^2 \cdot \frac{dy}{dx} \right) = 0 \]

3. Solve for \( \frac{dy}{dx} \implies \)

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

Example: Implicit Differentiation

Example: \( \sin(xy) + y^3 - 2x^3y = 7y^2 \), find \( y' \)

① \( \frac{d}{dx} \) on both sides

\[ \frac{d}{dx} \left[ \sin(xy) + y^3 - 2x^3y \right] = \frac{d}{dx}(7y^2) \] \[ \underbrace{\frac{d}{dx} \sin(xy)}_{\text{Chain Rule}} + \underbrace{\frac{d}{dx}(y^3)}_{\text{Chain Rule}} - \underbrace{\frac{d}{dx}(2x^3y)}_{\text{Product Rule}} = 7 \cdot \underbrace{\frac{d}{dx}(y^2)}_{\text{Chain Rule}} \]

Differentiating each term:

• \( \cos(xy) \cdot \underbrace{\frac{d}{dx}(xy)}_{\substack{\text{Product Rule} \\ 1 \cdot y + x\frac{dy}{dx}}} \)
• \( 3y^2 \frac{dy}{dx} \)
• \( 6x^2y + 2x^3 \frac{dy}{dx} \)
• \( 7 \cdot 2y \cdot \frac{dy}{dx} \)

\[ \cos(xy)\left[ y + x\left(\frac{dy}{dx}\right) \right] + 3y^2\left(\frac{dy}{dx}\right) - 6x^2y - 2x^3\left(\frac{dy}{dx}\right) = 14y\left(\frac{dy}{dx}\right) \]

Grouping and factoring out \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} \left[ x\cos(xy) + 3y^2 - 2x^3 - 14y \right] = -y\cos(xy) + 6x^2y \] \[ y' = (-y\cos(xy) + 6x^2y) / (x\cos(xy) + 3y^2 - 2x^3 - 14y) \]

Example: \( e^y = x \sin y \), find \( y' \)

\[ \frac{d}{dx} \left[ e^{y(x)} \right] = \frac{d}{dx} \left[ x \sin(y(x)) \right] \] \[ e^{y(x)} \cdot \frac{dy}{dx} = \frac{dx}{dx} \sin(y(x)) + x \cdot \frac{d}{dx} \sin(y(x)) \] \[ e^y \cdot y' = \sin y + x \cdot \cos(y) \cdot \frac{dy}{dx} = \sin y + x (\cos y) y' \] \[ y' \left[ e^y - x \cos y \right] = \sin y \] \[ y' = \frac{\sin y}{e^y - x \cos y} \]

\( e^y = x \sin y \), find \( y'' \)

found \( y' = \frac{dy}{dx} = \frac{\sin y}{e^y - x \cos y} \)

to find \( y'' \) take \( \frac{d}{dx} \) on both sides

\[ y'' = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left[\frac{\sin y}{e^y - x \cos y}\right] \] \[ = \frac{\frac{d}{dx}[\sin y](e^y - x \cos y) - \sin y\left[\frac{d}{dx}(e^y - x \cos y)\right]}{(e^y - x \cos y)^2} \]

\[ \frac{d}{dx}\sin y = \cos y \cdot y' \]

\[ \frac{d}{dx}(e^y - x \cos y) = e^y \cdot y' - \left[\frac{dx}{dx}\cos y + x \frac{d}{dx}\cos y\right] = e^y \cdot y' - [\cos y - x \sin y \cdot y'] \]

\[ y'' = \frac{(\cos y \cdot y')(e^y - x\cos y) - \sin y [e^y y' - \cos y + x\sin y \cdot y']}{(e^y - x\cos y)^2} \]

Replace:

\[ y' = \frac{\sin y}{e^y - x\cos y} \] \[ y'' = \frac{(\cos y)(\sin y) - \sin y \left[ e^y \cdot \frac{\sin y}{e^y - x\cos y} - \cos y + x\sin y \cdot \frac{\sin y}{e^y - x\cos y} \right]}{(e^y - x\cos y)^2} \]