Differentiating Inverse Functions - AP Calculus AB

  1. Find (f−1)′(2)(f^{-1})'(2) if f(x)=exf(x) = e^x and f(a)=2f(a) = 2.
    12\frac{1}{2}. Since f(ln⁡2)=2f(\ln 2) = 2 and f′(ln⁡2)=2f'(\ln 2) = 2, so (f−1)′(2)=12(f^{-1})'(2) = \frac{1}{2}.
  2. Find the derivative of y=arccos(5x)y = \text{arccos}(5x) at x=0x = 0.
    −5-5. Using chain rule: ddx[arccos⁡(5x)]=−51−(5x)2\frac{d}{dx}[\arccos(5x)] = -\frac{5}{\sqrt{1-(5x)^2}}, at x=0x=0 gives −5-5.
  3. Which inverse function has a derivative of −11+x2-\frac{1}{1+x^2}?
    arccot(x)\text{arccot}(x). Inverse cotangent has derivative −11+x2-\frac{1}{1+x^2}.
  4. Find the derivative of y=arcsin(2x)y = \text{arcsin}(2x) at x=0x = 0.
    22. Using chain rule: ddx[arcsin⁡(2x)]=21−(2x)2\frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1-(2x)^2}}, at x=0x=0 gives 22.
  5. Find (f−1)′(b)(f^{-1})'(b) if f(x)=x3+xf(x) = x^3 + x and f(a)=bf(a) = b.
    13a2+1\frac{1}{3a^2+1}. Using inverse function theorem: (f−1)′(b)=1f′(a)=13a2+1(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{3a^2+1}.
  6. Which inverse function has a derivative of 11+x2\frac{1}{1+x^2}?
    arctan(x)\text{arctan}(x). Inverse tangent has derivative 11+x2\frac{1}{1+x^2}.
  7. What is the domain of y=arccos(x)y = \text{arccos}(x)?
    [−1,1][-1, 1]. Domain restricted to values where −1≤x≤1-1 \leq x \leq 1.
  8. What is the derivative of y=arctan(x)y = \text{arctan}(x)?
    11+x2\frac{1}{1+x^2}. Standard derivative formula for inverse tangent function.
  9. Determine if y=arcsec(x)y = \text{arcsec}(x) is differentiable at x=0.5x = 0.5.
    No, y=arcsec(x)y = \text{arcsec}(x) is not differentiable at x=0.5x = 0.5. Domain of \arcsec\arcsec is ∣x∣≥1|x| \geq 1, so x=0.5x = 0.5 is outside domain.
  10. What is the derivative of y=arccot(x)y = \text{arccot}(x)?
    −11+x2-\frac{1}{1+x^2}. Standard derivative formula for inverse cotangent function.
  11. Find the derivative of y=arccot(6x)y = \text{arccot}(6x) at x=0x = 0.
    −6-6. Using chain rule: ddx[\arccot(6x)]=−61+(6x)2\frac{d}{dx}[\arccot(6x)] = -\frac{6}{1+(6x)^2}, at x=0x=0 gives −6-6.
  12. Determine if y=arcsin(x)y = \text{arcsin}(x) is differentiable at x=1x = 1.
    No, y=arcsin(x)y = \text{arcsin}(x) is not differentiable at x=1x = 1. At domain boundary, derivative is undefined.
  13. Find the derivative of y=\arccot(6x)y = \arccot(6x) at x=0x = 0.
    −6-6. Using chain rule: ddx[\arccot(6x)]=−61+(6x)2\frac{d}{dx}[\arccot(6x)] = -\frac{6}{1+(6x)^2}, at x=0x=0 gives −6-6.
  14. What is the domain of y=arcsin(x)y = \text{arcsin}(x)?
    [−1,1][-1, 1]. Domain restricted to values where −1≤x≤1-1 \leq x \leq 1.
  15. Find the derivative of y=arctan(3x)y = \text{arctan}(3x) at x=0x = 0.
    33. Using chain rule: ddx[arctan⁡(3x)]=31+(3x)2\frac{d}{dx}[\arctan(3x)] = \frac{3}{1+(3x)^2}, at x=0x=0 gives 33.
  16. State the formula for the derivative of an inverse function.
    (f−1)′(b)=1f′(a)(f^{-1})'(b) = \frac{1}{f'(a)} where f(a)=bf(a) = b. The inverse function derivative theorem.

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