Find the derivative of the inverse of the following functions using the inverse function rule.
f:x\longmapsto3x-1
The derivative of the inverse of a function f\left(x\right) is:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
The inverse of a function can be found by swapping x and f\left(x\right) and solving for f\left(x\right) :
f\left(x\right) = 3x - 1
x = 3 \cdot f^{-1}\left(x\right) - 1
3 \cdot f^{-1}\left(x\right) = x + 1
f^{-1}\left(x\right) = \dfrac{1}{3}x + \dfrac{1}{3}
This function is invertible.
The derivative of f\left(x\right) is:
\dfrac{d}{dx}f\left(x\right) = \dfrac{d}{dx}\left(3x - 1\right)
f'\left(x\right) = \dfrac{d}{dx}\left(3x\right) - \dfrac{d}{dx}\left(1\right)
f'\left(x\right) = 3 - 0
f'\left(x\right) = 3
Now the formula for the derivative of the inverse of f\left(x\right) can be used:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
\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}{3}
f : x \longmapsto \dfrac{2}{3}x^2 + 5 on x \geq 5
The derivative of the inverse of a function f\left(x\right) is:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
The inverse of a function can be found by swapping x and f\left(x\right) and solving for f\left(x\right) :
f\left(x\right) = \dfrac{2}{3}x^2 + 5
x = \dfrac{2}{3}\left(f^{-1}\left(x\right)\right)^2 + 5
\dfrac{2}{3}\left(f^{-1}\left(x\right)\right)^2 = x - 5
\left(f^{-1}\left(x\right)\right)^2 = \dfrac{3}{2}\left(x - 5\right)
Only the positive root is kept because f is defined for x\geq5 :
f^{-1}\left(x\right) = \sqrt{\dfrac{3}{2}\left(x - 5\right)}
This function is invertible and the inverse function has the domain x \geq 5.
The derivative of f\left(x\right) is:
\dfrac{d}{dx}f\left(x\right) = \dfrac{d}{dx}\left(\dfrac{2}{3}x^2 + 5\right)
f'\left(x\right) = \dfrac{2}{3} \cdot 2x = \dfrac{4}{3}x
Now the formula for the derivative of the inverse of f\left(x\right) can be used:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{\dfrac{4}{3}\sqrt{\dfrac{3}{2}\left(x - 5\right)}}
The function \left[f^{-1}\right]'\left(x\right) is undefined at x = 5. f^{-1}\left(x\right) exists only if the domain x \geq 5 and the function f^{-1}\left(x\right) is differentiable when x \gt 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 : x \longmapsto 3\sqrt{x}
The derivative of the inverse of a function f\left(x\right) is:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
The inverse of a function can be found by swapping x and f\left(x\right) and solving for f\left(x\right) :
f\left(x\right) = 3\sqrt{x}
x = 3\sqrt{f^{-1}\left(x\right)}
x^2 = 9 \cdot f^{-1}\left(x\right)
f^{-1}\left(x\right) = \dfrac{x^2}{9}
This function is invertible. Since the range of f\left(x\right) is f\left(x\right) \geq 0, the domain of f^{-1}\left(x\right) is x \geq 0. Since the domain of f^{-1}\left(x\right) is x \geq 0, it is dicontinuous at x = 0 and therefore differentiable when x \gt 0.
The derivative of f\left(x\right) is:
\dfrac{d}{dx}f\left(x\right) = \dfrac{d}{dx}\left(3\sqrt{x}\right)
f'\left(x\right) = 3 \cdot \dfrac{d}{dx}x^{\frac{1}{2}}
f'\left(x\right) = 3 \cdot \dfrac{1}{2} x^{\frac{1}{2} - 1}
f'\left(x\right) = 3 \cdot \dfrac{1}{2} x^{\frac{-1}{2}}
f'\left(x\right) = \dfrac{3}{2\sqrt{x}}
The domain for f'\left(x\right) is x \gt 0.
Now the formula for the derivative of the inverse of f\left(x\right) can be used:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{\dfrac{3}{2\sqrt{\dfrac{x^2}{9}}}}
\left[f^{-1}\right]'\left(x\right) = \dfrac{2}{3}\sqrt{\dfrac{x^2}{9}}
\left[f^{-1}\right]'\left(x\right) = \dfrac{2}{3} \cdot \dfrac{x}{3}
\left[f^{-1}\right]'\left(x\right) = \dfrac{2x}{9}
While \left[f^{-1}\right]' exists on x \geq 0, it is not differentiable when x = 0 because the domain for f'\left(x\right) is x \gt 0.
f^{-1} is differentiable for any x \gt 0 and \left[f^{-1}\right]'\left(x\right) = \dfrac{2x}{9}
f : x \longmapsto -2x^3 + 8
The derivative of the inverse of a function f\left(x\right) is:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
The inverse of a function can be found by swapping x and f\left(x\right) and solving for f\left(x\right) :
f\left(x\right) = -2x^3 + 8
x = -2 \cdot \left(f^{-1}\left(x\right)\right)^3 + 8
-2 \cdot \left(f^{-1}\left(x\right)\right)^3 = x - 8
\left(f^{-1}\left(x\right)\right)^3 = \dfrac{8 - x}{2}
f^{-1}\left(x\right) = \left(\dfrac{-x + 8}{2}\right)^\dfrac{1}{3}
This function is invertible and f^{-1}\left(x\right) exists on the domain \dfrac{8 - x}{2} \geq 0, or x \leq 8. The inverse function f^{-1}\left(x\right) is therefore discontinuous at x = 8 and differentiable only on x \lt 8.
The derivative of f\left(x\right) is:
\dfrac{d}{dx}f\left(x\right) = \dfrac{d}{dx}\left(-2x^3 + 8\right)
f'\left(x\right) = -6x^2
Now the formula for the derivative of the inverse of f\left(x\right) can be used:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{-6\left({\left(\frac{-x + 8}{2}\right)^\frac{1}{3}}\right)^2}
\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 \lt 8 and \left[f^{-1}\right]'\left(x\right) = \dfrac{-1}{6\left(\dfrac{8-x}{2}\right)^\dfrac{2}{3}}.
f : x \longmapsto \dfrac{6}{x^2} + 3 on x \gt 3
The derivative of the inverse of a function f\left(x\right) is:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
The inverse of a function can be found by swapping x and f\left(x\right) and solving for f\left(x\right) :
f\left(x\right) = \dfrac{6}{x^2} + 3
x = \dfrac{6}{\left[f^{-1}\left(x\right)\right]^2} + 3
x - 3 = \dfrac{6}{\left[f^{-1}\left(x\right)\right]^2}
\left[f^{-1}\left(x\right)\right]^2\left(x - 3\right) = 6
\left[f^{-1}\left(x\right)\right]^2 = \dfrac{6}{x-3}
Only the positive root is kept because f is defined for x \gt 3 :
f^{-1}\left(x\right) = \left(\dfrac{6}{x-3}\right)^{\frac{1}{2}}
This function is invertible on the domain x \gt 3.
The derivative of f\left(x\right) is:
\dfrac{d}{dx}f\left(x\right) = \dfrac{d}{dx}\left(\dfrac{6}{x^2} + 3\right)
f'\left(x\right) = 6 \cdot \dfrac{d}{dx}\left(x^{-2}\right) + \dfrac{d}{dx}\left(3\right)
f'\left(x\right) = 6 \cdot -2x^{-3}
f'\left(x\right) = \dfrac{-12}{x^3}
Now the formula for the derivative of the inverse of f\left(x\right) can be used:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{\dfrac{-12}{\left({\left(\frac{6}{x-3}\right)^{\frac{1}{2}}}\right)^{3}}}
\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 : x \longmapsto 2 \cdot e^x
The derivative of the inverse of a function f\left(x\right) is:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
The inverse of a function can be found by swapping x and f\left(x\right) and solving for f\left(x\right) :
f\left(x\right) = 2 \cdot e^x
x = 2 \cdot e^{f^{-1}\left(x\right)}
\ln\left(x\right) = \ln\left(2e{f^{-1}\left(x\right)}\right)
\ln\left(x\right) = \ln\left(2\right) + \ln\left(e{f^{-1}\left(x\right)}\right)
\ln\left(x\right) - \ln\left(2\right) = f^{-1}\left(x\right)
This function is invertible with the domain x \gt 0
The derivative of f\left(x\right) is:
\dfrac{d}{dx}f\left(x\right) = \dfrac{d}{dx}\left(2 \cdot e^x \right)
f'\left(x\right) = 2 \cdot e^x
Now the formula for the derivative of the inverse of f\left(x\right) can be used:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{2e^{\ln\left(x\right) - \ln\left(2\right)}}
\left[f^{-1}\right]'\left(x\right) = \dfrac{e^{\ln\left(2\right)}}{2e^{\ln\left(x\right)}}
\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{1}{x}
f : x \longmapsto 4 \cdot x^2 on -\infty \lt x \lt \infty
The derivative of the inverse of a function f\left(x\right) is:
\left[f^{-1}\right]'\left(x\right) = \dfrac{1}{f'\left(f^{-1}\left(x\right)\right)}
The inverse of a function can be found by swapping x and f\left(x\right) and solving for f\left(x\right) :
f\left(x\right) = 4 \cdot x^2
x = 4 \cdot \left[{f^{-1}\left(x\right)}\right]^2
\left[{f^{-1}\left(x\right)}\right]^2 = \dfrac{x}{4}
f^{-1}\left(x\right) = \dfrac{\sqrt{x}}{2}, or f^{-1}\left(x\right) = -\dfrac{\sqrt{x}}{2}
f^{-1}\left(x\right) has more than one unique value for a single given value of x and is therefore not a function. f\left(x\right) is therefore not invertible.
f^{-1} does not exist.