Notes on Differentiating Inverse Functions and Related Rates

Derivatives of Inverse Functions

  • Inverse trigonometric functions have specific differentiation rules derived from implicit differentiation.

Example: Derivative of Arc Sine

  • To find ( \frac{dy}{dx} ) when ( y = \arcsin(x) ):

  • Recall: ( \arcsin(x) ) is the inverse of ( \sin(x) ).

  • To differentiate, take the sine of both sides:

    • ( \sin(y) = x )

  • Differentiate using implicit differentiation:

    • ( \frac{d}{dx} \sin(y) = \frac{d}{dx}x )
    • This results in: ( \cos(y) \cdot \frac{dy}{dx} = 1 )

  • Solving for ( \frac{dy}{dx} ):

    • ( \frac{dy}{dx} = \frac{1}{\cos(y)} )

  • To express in terms of ( x ):

  • Use the identity: ( \sin^2(y) + \cos^2(y) = 1 )

  • Therefore, ( \cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - x^2} )

  • Final Result: ( \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} )

Other Inverse Trig Derivatives

  • Arc Cosine: ( \frac{d}{dx} \arccos(x) = -\frac{1}{\sqrt{1 - x^2}} )
  • Arc Tangent: ( \frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2} )
  • Arc Secant: ( \frac{d}{dx} \text{arccsc(x)} = -\frac{1}{|x| \sqrt{x^2 - 1}} )
  • Arc Cotangent: ( \frac{d}{dx} \text{arccot(x)} = -\frac{1}{1 + x^2} )

Example: Chain Rule with Inverse Functions

  • Finding ( \frac{d}{dx} \arctan(x^2) ):
  • Apply chain rule since it's ( \arctan(f(x)) ):
    • ( \frac{d}{dx} \arctan(x^2) = \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2) )
  • Result:
    • ( = \frac{2x}{1 + x^4} )

Example: Derivative of Arc Secant

  • Finding ( \frac{d}{dx} \text{arcsec}(e^x) ):
  • Again, use chain rule:
    • ( \frac{d}{dx} \text{arcsec}(e^x) = \frac{1}{|e^x|\sqrt{(e^x)^2 - 1}} \cdot \frac{d}{dx}(e^x) )
  • Final Result:
    • Simplifies to ( \frac{e^x}{\sqrt{e^{2x} - 1}} )

Applying Implicit Differentiation

  • Recognize that implicit differentiation underpins the rules of derivatives of inverse functions.
  • For more complex functions, derive using implicit differentiation as shown in examples.

Practice Problem

What is the slope of the tangent line to the curve defined by inverse functions?

  • To find the slope at a specific point (e.g., ( (1,1) )), substitute the point's coordinates into the derived expression for ( \frac{dy}{dx} ).
  • Ensure the chosen point is within the domain of the function for meaningful values.

Upcoming Topics

  • Related Rates:
  • Emphasis on using implicit differentiation for word problems involving rates of change.
  • This will be covered in the next session.