8.2 Applications of implicit differentiation

Inverse trig functions and implicit differentiation are powerful tools for solving complex calculus problems. They let us find derivatives of tricky functions and tackle real-world scenarios involving changing rates.

These techniques open doors to optimization and related rates problems. By using implicit differentiation, we can analyze situations where multiple variables change over time, helping us solve practical math challenges in various fields.

Inverse Trigonometric Functions and Implicit Differentiation

Derivatives of inverse trigonometric functions

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• Inverse trigonometric functions defined implicitly enable differentiation using the implicit differentiation technique
  • $y = \arcsin(x)$ implicitly defined as $x = \sin(y)$ ($y$ is the angle whose sine is $x$)
  • $y = \arccos(x)$ implicitly defined as $x = \cos(y)$ ($y$ is the angle whose cosine is $x$)
  • $y = \arctan(x)$ implicitly defined as $x = \tan(y)$ ($y$ is the angle whose tangent is $x$)
• Differentiating both sides of the implicit equation with respect to $x$ yields the derivative of the inverse trigonometric function
  • For $y = \arcsin(x)$, differentiate $x = \sin(y)$:
    • $1 = \cos(y) \cdot \frac{dy}{dx}$ (using the chain rule)
    • $\frac{dy}{dx} = \frac{1}{\cos(y)}$ (solving for $\frac{dy}{dx}$)
    • $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$ (substituting $y = \arcsin(x)$ and simplifying)
  • For $y = \arccos(x)$, differentiate $x = \cos(y)$:
    • $1 = -\sin(y) \cdot \frac{dy}{dx}$ (using the chain rule and the derivative of cosine)
    • $\frac{dy}{dx} = -\frac{1}{\sin(y)}$ (solving for $\frac{dy}{dx}$)
    • $\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}$ (substituting $y = \arccos(x)$ and simplifying)
  • For $y = \arctan(x)$, differentiate $x = \tan(y)$:
    • $1 = \sec^2(y) \cdot \frac{dy}{dx}$ (using the chain rule and the derivative of tangent)
    • $\frac{dy}{dx} = \frac{1}{\sec^2(y)}$ (solving for $\frac{dy}{dx}$)
    • $\frac{dy}{dx} = \frac{1}{1+x^2}$ (substituting $y = \arctan(x)$ and simplifying)

Related Rates and Optimization

Related rates through implicit differentiation

• Related rates problems involve multiple variables changing with respect to time, related by an equation
• Solving related rates problems:
  1. Identify variables and write an equation relating them
  2. Differentiate both sides of the equation with respect to time ($t$)
  3. Substitute known values and solve for the desired rate
• Example: A 10 ft ladder rests against a vertical wall. The bottom slides away at 1 ft/sec. How fast is the top sliding down when the bottom is 6 ft from the wall?
  • $x$: distance from bottom of ladder to wall, $y$: distance from top of ladder to ground
  • Pythagorean theorem: $x^2 + y^2 = 10^2$
  • Differentiate with respect to time: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
  • Substitute known values: $2(6)(1) + 2y\frac{dy}{dt} = 0$
  • Solve for $\frac{dy}{dt}$ when $x = 6$ (and $y = 8$): $\frac{dy}{dt} = -\frac{6}{8} = -0.75$ ft/sec

Optimization with implicit functions

• Optimization problems involve finding maximum or minimum values of a function subject to constraints
• For implicitly defined functions, use implicit differentiation to find critical points
• Example: Find points on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ farthest from and closest to the origin
  • Distance from point $(x, y)$ to origin: $d = \sqrt{x^2 + y^2}$
  • Substitute ellipse equation: $d = \sqrt{a^2(\frac{x^2}{a^2}) + b^2(\frac{y^2}{b^2})} = \sqrt{a^2 + b^2 - a^2(\frac{y^2}{b^2})}$
  • Find critical points by differentiating with respect to $y$ and setting equal to zero:
    • $\frac{d}{dy}(\sqrt{a^2 + b^2 - a^2(\frac{y^2}{b^2})}) = 0$
    • $\frac{-a^2(\frac{2y}{b^2})}{\sqrt{a^2 + b^2 - a^2(\frac{y^2}{b^2})}} = 0$
    • $y = 0$ (maximum) or $y = \pm b$ (minimum)
  • Points farthest from origin: $(0, \pm b)$, points closest to origin: $(\pm a, 0)$

Logarithmic Functions

Differentiation of logarithmic functions

• Logarithmic functions can be differentiated using implicit differentiation
• Natural logarithm function $\ln(x)$ implicitly defined as $e^y = x$
  • Differentiate both sides with respect to $x$: $e^y \cdot \frac{dy}{dx} = 1$
  • Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{e^y}$
  • Substitute $y = \ln(x)$ to get $\frac{d}{dx}(\ln(x)) = \frac{1}{x}$
• General logarithm function $\log_b(x)$ expressed in terms of natural logarithm: $\log_b(x) = \frac{\ln(x)}{\ln(b)}$
  • Differentiate using chain rule: $\frac{d}{dx}(\log_b(x)) = \frac{1}{\ln(b)} \cdot \frac{1}{x}$
• Logarithmic differentiation used for functions like $y = x^x$ or $y = (f(x))^{g(x)}$
  1. Take natural logarithm of both sides
  2. Differentiate using implicit differentiation
  • Example: Differentiate $y = x^x$
    • $\ln(y) = x\ln(x)$
    • $\frac{1}{y} \cdot \frac{dy}{dx} = \ln(x) + 1$ (using implicit differentiation and the chain rule)
    • $\frac{dy}{dx} = x^x(\ln(x) + 1)$ (solving for $\frac{dy}{dx}$ and substituting $y = x^x$)