Derivatives of inverse trigonometric functions:

• If \(y = \sin(x)\), then \(x = \arcsin(y)\), which means the inverse relationship allows us to find the angle
• The derivative of \(\arcsin(x)\) is \(\frac{1}{\sqrt{1-x^2}}\)

  True
• Match each basic differentiation rule with its formula:

  Power Rule ↔️ \(\frac{d}{dx}(x^n) = nx^{n-1}\)

  Chain Rule ↔️ \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)

  Quotient Rule ↔️ \(\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\)
• What is the derivative of \(\arccos(x)\)?

  \(\frac{-1}{\sqrt{1-x^2}}\)
• The derivative of \(\arccos(x)\) is \(\frac{-1}{\sqrt{1-x^2}}\), which is the negative of the derivative of \arcsin(x)
• If \(y = \sin(x)\), then \(x = \arcsin(y)\)


  True
• The derivatives of the inverse trigonometric functions are used in calculus for finding slopes of tangent lines.
• What is the derivative of \(\sin(x^2)\)?

  $\cos(x^{2}) \cdot 2x$
• What is the derivative of \(\arctan(x)\)?

  $\frac{1}{1 + x^{2}}$
• If \(y = \cos(x)\), then what is \(x\)?

  \(\arccos(y)\)
• What is the inverse of \(y = \tan(x)\)?

  \(\arctan(y)\)
• The derivative of `arcsin(x)` is \frac{1}{\sqrt{1-x^2}}
• What is the derivative of \arctan x</latex>?

  \frac{1}{1+x^2}
• The chain rule formula is $\frac{d}{dx}f(g(x)) =$$f'(g(x)) \cdot g'(x)$.


  True
• If `y = arctan(x+1)`, then `dy/dx =` \frac{1}{x^{2} +$2x +$ 2}.


  True
• Steps to apply the chain rule to inverse trigonometric functions

  1️⃣ Identify the inner and outer functions

  2️⃣ Find the derivatives of both functions

  3️⃣ Use the chain rule formula
• If \(y = \tan(x)\), then \(x = \arctan(y)\).


  True
• What is the derivative of $\arccos(x)$?


  $\frac{ - 1}{\sqrt{1 - x^{2}}}$
• The derivative of $\arctan x$ is $\frac{1}{1 + x^{2}}$.


  True
• What is the first step in applying the chain rule to differentiate inverse trigonometric functions?

  Identify inner and outer functions
• Match the function with its derivative:

  \(\arcsin(x^2)\) ↔️ \(\frac{2x}{\sqrt{1 - x^4}}\)

  \(\arccos(2x)\) ↔️ \(\frac{-2}{\sqrt{1 - 4x^2}}\)

  \(\arctan(x+1)\) ↔️ \(\frac{1}{x^2 + 2x + 2}\)
• Steps to differentiate composite functions involving inverse trigonometric functions

  1️⃣ Identify the inner and outer functions

  2️⃣ Find the derivatives of both the inner and outer functions

  3️⃣ Use the chain rule formula: \(\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\)
• For \(y = \arccos(2x)\), the derivative is \frac{-2}{\sqrt{1 - 4x^2}}
• For \(y = \arcsin(x^2)\), the derivative is \frac{2x}{\sqrt{1 - x^4}}
• Systematically applying the chain rule is essential for finding derivatives of inverse trigonometric functions.

  True
• What are the three most common inverse trigonometric functions?

  arcsin, arccos, arctan
• The derivative of \(\arctan(x)\) is \frac{1}{1+x^2}
• What is the derivative of \(x^2 \cdot \sin(x)\) using the product rule?

  2x\sin(x) + x^2\cos(x)
• Steps to differentiate inverse trigonometric functions using the chain rule:

  1️⃣ Apply the chain rule formula

  2️⃣ Differentiate the outer function

  3️⃣ Differentiate the inner function

  4️⃣ Multiply the results
• What is the derivative of \(\arcsin(x)\)?

  $\frac{1}{\sqrt{1 - x^{2}}}$
• Match each inverse trigonometric function with its inverse relationship:

  arcsin ↔️ \(y = \sin(x)\)

  arccos ↔️ \(y = \cos(x)\)

  arctan ↔️ \(y = \tan(x)\)
• The power rule states that $\frac{d}{dx}(x^{n}) =$$nx^{n - 1}$


  True
• Mastering basic differentiation rules is essential for complex function differentiation.


  True
• The derivatives of inverse trigonometric functions are used for solving optimization problems.


  True
• The inverse trigonometric functions are denoted with the prefix arc.
• What prefix is used to denote inverse trigonometric functions?

  arc
• What is the derivative of `arccos(x)`?

  \frac{-1}{\sqrt{1-x^2}}
• If `f(x) = arcsin(3x)`, then `f'(x) =` \frac{3}{\sqrt{1 - (3x)^2}}
• If `y = arcsin(x^2)`, then `dy/dx =` \frac{2x}{\sqrt{1 - x^4}}
• What is the derivative of $y =$$\arccos(2x)$?


  $\frac{ - 2}{\sqrt{1 - 4x^{2}}}$