Video Summary - (심화수학) 역삼각함수의 도함수 (1)

Main ideas / lessons

The lesson builds on what you already know about inverse trigonometric functions by focusing on how to differentiate them—specifically, how to find derivatives of inverse trig functions.
It starts by reviewing key composition properties of inverse trig functions:
- Composing an inverse trig function with its original trig function returns the input, but only when the input is restricted to the domain where the inverse is defined.
It then uses those properties to derive derivative formulas by:
- Writing a relationship between an inverse trig function and the original trig function,
- Differentiating both sides implicitly,
- Solving for ( \frac{dy}{dx} ),
- Converting any remaining expression in terms of (y) into an expression in terms of (x),
- Enforcing domain restrictions, especially when:
  - Dividing by expressions that must be nonzero, and
  - Respecting the inverse function’s definition interval.

Start with an inverse trig definition, typically of the form:
- (y = \arcsin(x)), (y = \arccos(x)), or (y = \arctan(x))
Rewrite into trig form. For example:
- (y=\arcsin(x)\Rightarrow x=\sin y)
Differentiate both sides with respect to (x):
- Use the chain rule (differentiate the trig function of (y) and multiply by ( \frac{dy}{dx} ))
Solve the resulting equation for ( \frac{dy}{dx} ).
Replace trig expressions involving (y) with equivalent expressions involving (x):
- Often using identities like ( \sin^2 y + \cos^2 y = 1 )
Determine and enforce domain restrictions:
- Ensure denominators are not zero (which may exclude endpoint values),
- Ensure the inverse function is used only on its standard input interval.

Begin with:
- ( y=\arcsin(x) \Rightarrow x=\sin y )
Differentiate implicitly and solve for ( \frac{dy}{dx} ).
Convert using:
- ( \cos^2 y = 1-\sin^2 y = 1-x^2 )
Choose the correct sign:
- Because of the restricted range used for (\arcsin), the derivation emphasizes (\cos y \ge 0), so:
  - ( \cos y = \sqrt{1-x^2} )
Final derivative:
- [ \frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}} ]
Domain/exclusion emphasized:
- At endpoints (x=\pm 1), the denominator becomes (0), so the derivative is not valid there.
- The derivation also notes that the step involving division requires (\cos y \ne 0).

Use:
- ( y=\arccos(x) \Rightarrow x=\cos y )
Differentiate implicitly:
- Derivative of (\cos y) gives (-\sin y\cdot \frac{dy}{dx}).
Solve for ( \frac{dy}{dx} ) and ensure division is valid (requires (\sin y \ne 0)).
Convert using:
- ( \sin^2 y = 1-\cos^2 y = 1-x^2 )
Sign reasoning using the inverse-cosine range:
- The derivation argues that within the standard (\arccos) range, (\sin y \ge 0), so:
  - ( \sin y = \sqrt{1-x^2} )
Final derivative:
- [ \frac{d}{dx}\arccos(x) = -\frac{1}{\sqrt{1-x^2}} ]
Domain/exclusion emphasized:
- (x=\pm 1) makes the denominator (0).
- The intermediate step also divides by (\sin y), which becomes (0) at the corresponding endpoint cases.

Use:
- ( y=\arctan(x) \Rightarrow x=\tan y )
Differentiate implicitly:
- [ \frac{d}{dx}(\tan y)=\sec^2 y\cdot \frac{dy}{dx} ]
Solve:
- [ \sec^2 y\cdot \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx}=\frac{1}{\sec^2 y} ]
Convert (\sec^2 y) in terms of (x):
- Since (\tan y = x),
- [ \sec^2 y = 1+\tan^2 y = 1+x^2 ]
Final derivative:
- [ \frac{d}{dx}\arctan(x) = \frac{1}{1+x^2} ]
Domain note emphasized:
- Unlike (\arcsin) and (\arccos), (\arctan) has no real (x) endpoint singularities in the derivative formula.

No specific named speaker or on-screen identity is provided in the subtitles.
The content appears to be delivered by a single instructor/lecturer (speaker not identified by name).