3.4 Differentiating Inverse Trigonometric Functions

    Cards (51)

    • What is the primary purpose of inverse trigonometric functions?
      Find corresponding angles
    • The inverse trigonometric function Arcsin finds the angle whose sine is xx.
    • What is the domain of the arcsin(x) function?
      [1,1][ - 1, 1]
    • The range of the arctan(x) function is ( - π / 2, π / 2).
    • arcsin(1 / 2) is equal to π / 6.
    • The derivative of arcsin(x)arcsin(x) is 11x2\frac{1}{\sqrt{1 - x^{2}}}.
    • The derivative of arccos(x)arccos(x) is 11x2\frac{ - 1}{\sqrt{1 - x^{2}}}.
    • The derivative of arctan(x)arctan(x) is 11+x2\frac{1}{1 + x^{2}}.
    • Match the inverse trigonometric function with its derivative:
      arcsin(x)arcsin(x) ↔️ 11x2\frac{1}{\sqrt{1 - x^{2}}}
      arccos(x)arccos(x) ↔️ 11x2\frac{ - 1}{\sqrt{1 - x^{2}}}
      arctan(x)arctan(x) ↔️ 11+x2\frac{1}{1 + x^{2}}
    • The chain rule formula is ddx[f(g(x))]=\frac{d}{dx}[f(g(x))] =f(g(x))g(x) f'(g(x)) \cdot g'(x).
    • The derivative of arcsin(3x)arcsin(3x) is 319x2\frac{3}{\sqrt{1 - 9x^{2}}}.
    • What is the derivative of arctan(x2)arctan(x^{2})?

      2x1+x4\frac{2x}{1 + x^{4}}
    • The derivative of arccos(x)arccos(\sqrt{x}) is 12x1x\frac{ - 1}{2\sqrt{x}\sqrt{1 - x}}.
    • Arcsin(x)Arcsin(x) finds the angle whose sine is xx.
    • Why are the domains and ranges of inverse trigonometric functions restricted?
      To ensure unique values
    • The range of arcsin(x) is [π/2,π/2][ - π / 2, π / 2]
    • The range of arccos(x) is [0,π][0, π]
    • What is the domain of arctan(x)?
      (,)( - ∞, ∞)
    • arcsin(1/2)=arcsin(1 / 2) = π / 6</latex>
    • arccos(0)=arccos(0) =π/2 π / 2
    • What is the value of arctan(1)arctan(1)?

      π/4π / 4
    • Order the inverse trigonometric functions by the complexity of their derivatives:
      1️⃣ arctan(x)arctan(x)
      2️⃣ arcsin(x)arcsin(x)
      3️⃣ arccos(x)arccos(x)
    • What is the derivative of arcsin(x)arcsin(x)?

      11x2\frac{1}{\sqrt{1 - x^{2}}}
    • The derivative of arccos(x)arccos(x) is 11x2\frac{1}{\sqrt{1 - x^{2}}}
      False
    • The derivative of arctan(x)arctan(x) is 11+x2\frac{1}{1 + x^{2}}
    • What is the derivative of arcsin(x)</latex> if y=y =arcsin(x) arcsin(x)?

      11x2\frac{1}{\sqrt{1 - x^{2}}}
    • The derivative of arccos(x)arccos(x) is negative.
    • Match the composite inverse trigonometric function with its correct derivative:
      arcsin(g(x))arcsin(g(x)) ↔️ g(x)1g(x)2\frac{g'(x)}{\sqrt{1 - g(x)^{2}}}
      arccos(g(x))arccos(g(x)) ↔️ g(x)1g(x)2\frac{ - g'(x)}{\sqrt{1 - g(x)^{2}}}
      arctan(g(x))arctan(g(x)) ↔️ g(x)1+g(x)2\frac{g'(x)}{1 + g(x)^{2}}
    • The derivative of arcsin(3x)arcsin(3x) is 319x2\frac{3}{\sqrt{1 - 9x^{2}}}
    • The derivative of arctan(x2)arctan(x^{2}) is 2x1+x4\frac{2x}{1 + x^{4}}
    • What is the derivative of arccos(x)arccos(\sqrt{x})?

      12x1x\frac{ - 1}{2\sqrt{x}\sqrt{1 - x}}
    • Steps to differentiate an implicit function involving inverse trigonometric functions:
      1️⃣ Differentiate both sides
      2️⃣ Use the chain rule
      3️⃣ Solve for dydx\frac{dy}{dx}
    • Given arctan(y)+arctan(y) +x= x =1 1, what is dydx\frac{dy}{dx}?

      (1+y2)- (1 + y^{2})
    • To differentiate inverse trigonometric functions, the chain rule is always necessary.
    • The derivative of arctan(x2)arctan(x^{2}) is 2x1+x4\frac{2x}{1 + x^{4}}
    • The derivative of arccos(x)arccos(x) is - 1 / \sqrt{1 - x^{2}}
    • The derivative of arctan(x)arctan(x) is 1 / (1 + x^{2})
    • Steps to find the derivative of y=y =arctan(x2) arctan(x^{2})
      1️⃣ Identify the inner function: g(x)=g(x) =x2 x^{2}
      2️⃣ Find its derivative: g(x)=g'(x) =2x 2x
      3️⃣ Apply the chain rule: dydx=\frac{dy}{dx} =11+(x2)22x \frac{1}{1 + (x^{2})^{2}} \cdot 2x
      4️⃣ Simplify the result: dydx=\frac{dy}{dx} =2x1+x4 \frac{2x}{1 + x^{4}}
    • The derivative of y=y =arctan(x2) arctan(x^{2}) is \frac{2x}{1 + x^{4}}</latex>
    • The inverse trigonometric functions have a restricted domain and range
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