Find the derivative of the inverse of a function - High school Calculus Exercises

Find the derivative of the inverse of the following functions using the inverse function rule.

f:x\longmapsto3x-1

f^{-1} is differentiable for any x \in \mathbb{R} and \left[f^{-1}\right]'\left(x\right)=\dfrac{1}{3}

f^{-1} is differentiable for any x \in \mathbb{R} and \left[f^{-1}\right]'\left(x\right)= \dfrac{1}{x}

f^{-1} is differentiable for any x \in \mathbb{R} and \left[f^{-1}\right]'\left(x\right)= \dfrac{3}{x}

f^{-1} is not differentiable

f : x \longmapsto \dfrac{2}{3}x^2 + 5 on x \geq 5

f^{-1} is differentiable for any x \gt 5 and \left[f^{-1}\right]'\left(x\right) = \dfrac{3}{4\sqrt{\dfrac{3}{2}\left(x - 5\right)}}

f^{-1} is differentiable for any x \geq 5 and \left[f^{-1}\right]'\left(x\right) = \dfrac{3}{4\sqrt{\dfrac{3}{2}\left(x - 5\right)}}

f^{-1} is differentiable for any x \gt 5 and \left[f^{-1}\right]'\left(x\right) = \dfrac{3}{4}\sqrt{\dfrac{3}{2}\left(x - 5\right)}

f^{-1} is not differentiable

f : x \longmapsto 3\sqrt{x}

f^{-1} is differentiable for any x \gt 0 and \left[f^{-1}\right]'\left(x\right) = \dfrac{2x}{9}

f^{-1} is differentiable for any x \gt 0 and \left[f^{-1}\right]'\left(x\right) = \dfrac{2x}{3}

f^{-1} is differentiable for any x \geq 0 and \left[f^{-1}\right]'\left(x\right) = \dfrac{2x}{3\sqrt{3}}

f^{-1} is not differentiable

f : x \longmapsto -2x^3 + 8

f^{-1} is differentiable for any x \lt 8 and \left[f^{-1}\right]'\left(x\right) = \dfrac{-1}{6\left(\dfrac{8-x}{2}\right)^\dfrac{2}{3}}

f^{-1} is differentiable for any x \in \mathbb{R} and \left[f^{-1}\right]'\left(x\right) = -x

f^{-1} is differentiable for any x \gt -8 and \left[f^{-1}\right]'\left(x\right) = \dfrac{-1}{6\left(\dfrac{-8+x}{2}\right)^\dfrac{2}{3} + 8}

f^{-1} is not differentiable

f : x \longmapsto \dfrac{6}{x^2} + 3 on x \gt 3

f^{-1} is differentiable for any x \gt 3 and \left[f^{-1}\right]'\left(x\right) = \dfrac{-\left(\dfrac{6}{x-3}\right)^\dfrac{3}{2}}{12}

f^{-1} is differentiable for any x \gt 3 and \left[f^{-1}\right]'\left(x\right) = \dfrac{-\left(\dfrac{6}{x-3}\right)^\dfrac{-3}{2}}{12}

f^{-1} is differentiable for any x \gt 3 and \left[f^{-1}\right]'\left(x\right) = \dfrac{1}{2\left(x-3\right)^{\frac{3}{2}}}

f^{-1} is not differentiable

f : x \longmapsto 2 \cdot e^x

f^{-1} is differentiable for any x \gt 0 and \left[f^{-1}\right]'\left(x\right) = \dfrac{1}{x}

f^{-1} is differentiable for any x \gt 0 and \left[f^{-1}\right]'\left(x\right) = \dfrac{e^{\ln\left(2\right)}}{2\left(\ln\left(x\right) - \ln\left(2\right)\right)e^{\ln\left(x\right)}}

f^{-1} is differentiable for any x \gt 0 and \left[f^{-1}\right]'\left(x\right) = \dfrac{e^{\ln\left(\frac{2}{x}\right)}}{2}

f^{-1} is not differentiable

f : x \longmapsto 4 \cdot x^2 on -\infty \lt x \lt \infty

f^{-1} does not exist

f^{-1} is differentiable for any x \gt 0 and \left[f^{-1}\right]'\left(x\right) = \dfrac{1}{4\sqrt{x}}

f^{-1} is differentiable for any x \gt 0 and \left[f^{-1}\right]'\left(x\right) = \dfrac{1}{2\sqrt{x}}

f^{-1} is differentiable for any x \geq 0 and \left[f^{-1}\right]'\left(x\right) = 4\sqrt{x}