The derivative of an inverse function allows us to find the slope of an inverse function without explicitly solving for the full inverse equation. The key formula is \(\left(f^{-1}\right)'(a)=\frac{1}{f'(f^{-1}(a))}\), which uses the derivative of the original function evaluated at the input that gives output \(a\). These problems focus on finding \(f^{-1}(a)\), evaluating \(f'(x)\), and then applying the inverse function derivative formula.
Derivative of the Inverse Function
\(\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’ \left( f^{-1}(a) \right)}\)
Practice Problems
\(\textbf{1)}\) Determine \(\left(f^{-1}\right)'(8)\) for \(f(x) = x^3\).
The answer is \(\displaystyle \left(f^{-1}\right)'(8)= \frac{1}{12}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’ \left( f^{-1}(a) \right)}\)
\(\text{Step 2: Fill in for }a=8\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(8)=\frac{1}{f’ \left( f^{-1}(8) \right)}\)
\(\text{Step 3: Solve for }f^{-1}(8)\)
\(\,\,\,\,\,\, 8=x^3 \)
\(\,\,\,\,\,\, x=2 \)
\(\,\,\,\,\,\, f^{-1}(8)=2\)
\(\text{Step 4: Solve for }f'(2)\)
\(\,\,\,\,\,\, f'(x)=3x^2 \)
\(\,\,\,\,\,\, f'(2)=3(2)^2=12\)
\(\text{Step 5: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(8)=\frac{1}{f'(2)}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(8)=\frac{1}{12}\)
\(\text{The answer is }\displaystyle\left(f^{-1}\right)'(8)=\frac{1}{12}\)
\(\textbf{2)}\) Determine \(\left(f^{-1}\right)'(6)\) for \(f(x) = 15-x^2, x\ge 0\).
The answer is \(\displaystyle \left(f^{-1}\right)'(6)=\, – \,\frac{1}{6}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’ \left( f^{-1}(a) \right)}\)
\(\text{Step 2: Fill in for }a=6\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(6)=\frac{1}{f’ \left( f^{-1}(6) \right)}\)
\(\text{Step 3: Solve for }f^{-1}(6)\)
\(\,\,\,\,\,\, 6=15-x^2 \)
\(\,\,\,\,\,\, x^2=9 \)
\(\,\,\,\,\,\, x=3 \left(\text{since the domain is } x\ge0\right)\)
\(\,\,\,\,\,\, f^{-1}(6)=3\)
\(\text{Step 4: Solve for }f'(3)\)
\(\,\,\,\,\,\, f'(x)=-2x \)
\(\,\,\,\,\,\, f'(3)=-6\)
\(\text{Step 5: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(6)=\frac{1}{f'(3)}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(6)=\frac{1}{-6}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(6)=\, -\,\frac{1}{6}\)
\(\textbf{3)}\) Determine \(\left(f^{-1}\right)'(4)\) for \(f(x) = \sqrt{x}\).
The answer is \(\displaystyle \left(f^{-1}\right)'(4)=8\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’ \left( f^{-1}(a) \right)}\)
\(\text{Step 2: Fill in for }a=4\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(4)=\frac{1}{f’ \left( f^{-1}(4) \right)}\)
\(\text{Step 3: Solve for }f^{-1}(4)\)
\(\,\,\,\,\,\, 4=\sqrt{x} \)
\(\,\,\,\,\,\, x=16 \)
\(\,\,\,\,\,\, f^{-1}(4)=16\)
\(\text{Step 4: Solve for }f'(16)\)
\(\,\,\,\,\,\, f'(x)=\frac{1}{2\sqrt{x}} \)
\(\,\,\,\,\,\, f'(16)=\frac{1}{2\sqrt{16}}=\frac{1}{8}\)
\(\text{Step 5: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(4)=\frac{1}{f'(16)}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(4)=\frac{1}{\frac{1}{8}}\)
\(\text{The answer is }\displaystyle\left(f^{-1}\right)'(4)=8\)
\(\textbf{4)}\) Determine \(\left(f^{-1}\right)'(9)\) for \(f(x) = x^2, x\ge 0\).
The answer is \(\displaystyle \left(f^{-1}\right)'(9)= \frac{1}{6}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’ \left( f^{-1}(a) \right)}\)
\(\text{Step 2: Fill in for }a=9\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(9)=\frac{1}{f’ \left( f^{-1}(9) \right)}\)
\(\text{Step 3: Solve for }f^{-1}(9)\)
\(\,\,\,\,\,\, 9=x^2 \)
\(\,\,\,\,\,\, x=3 \left(\text{since the domain is } x\ge0\right)\)
\(\,\,\,\,\,\, f^{-1}(9)=3\)
\(\text{Step 4: Solve for }f'(3)\)
\(\,\,\,\,\,\, f'(x)=2x \)
\(\,\,\,\,\,\, f'(3)=6\)
\(\text{Step 5: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(9)=\frac{1}{f'(3)}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(9)= \frac{1}{6}\)
\(\textbf{5)}\) Determine \(\left(f^{-1}\right)'(1)\) for \(f(x) = x^2-4x+4, x\ge 2\).
The answer is \(\displaystyle \left(f^{-1}\right)'(1)= \frac{1}{2}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’ \left( f^{-1}(a) \right)}\)
\(\text{Step 2: Fill in for }a=1\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(1)=\frac{1}{f’ \left( f^{-1}(1) \right)}\)
\(\text{Step 3: Solve for }f^{-1}(1)\)
\(\,\,\,\,\,\, 1=x^2-4x+4 \)
\(\,\,\,\,\,\, 0=x^2-4x+3 \)
\(\,\,\,\,\,\, 0=(x-1)(x-3)\)
\(\,\,\,\,\,\, x=3 \left(\text{since the domain is } x\ge2\right)\)
\(\,\,\,\,\,\, f^{-1}(1)=3\)
\(\text{Step 4: Solve for }f'(3)\)
\(\,\,\,\,\,\, f'(x)=2x-4 \)
\(\,\,\,\,\,\, f'(3)=2\)
\(\text{Step 5: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(1)=\frac{1}{f'(3)}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(1)= \frac{1}{2}\)
\(\textbf{6)}\) Determine \(\left(f^{-1}\right)'(27)\) for \(f(x)=x^3\).
The answer is \(\displaystyle \left(f^{-1}\right)'(27)=\frac{1}{27}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(27)\)
\(\,\,\,\,\,\,27=x^3\)
\(\,\,\,\,\,\,x=3\)
\(\,\,\,\,\,\,f^{-1}(27)=3\)
\(\text{Step 3: Find }f'(3)\)
\(\,\,\,\,\,\,f'(x)=3x^2\)
\(\,\,\,\,\,\,f'(3)=27\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(27)=\frac{1}{27}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(27)=\frac{1}{27}\)
\(\textbf{7)}\) Determine \(\left(f^{-1}\right)'(16)\) for \(f(x)=x^2,\;x\ge0\).
The answer is \(\displaystyle \left(f^{-1}\right)'(16)=\frac{1}{8}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(16)\)
\(\,\,\,\,\,\,16=x^2\)
\(\,\,\,\,\,\,x=4\left(\text{since }x\ge0\right)\)
\(\,\,\,\,\,\,f^{-1}(16)=4\)
\(\text{Step 3: Find }f'(4)\)
\(\,\,\,\,\,\,f'(x)=2x\)
\(\,\,\,\,\,\,f'(4)=8\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(16)=\frac{1}{8}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(16)=\frac{1}{8}\)
\(\textbf{8)}\) Determine \(\left(f^{-1}\right)'(25)\) for \(f(x)=\sqrt{x}\).
The answer is \(\displaystyle \left(f^{-1}\right)'(25)=50\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(25)\)
\(\,\,\,\,\,\,25=\sqrt{x}\)
\(\,\,\,\,\,\,x=625\)
\(\,\,\,\,\,\,f^{-1}(25)=625\)
\(\text{Step 3: Find }f'(625)\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x}}\)
\(\,\,\,\,\,\,f'(625)=\frac{1}{2(25)}=\frac{1}{50}\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(25)=\frac{1}{\frac{1}{50}}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(25)=50\)
\(\textbf{9)}\) Determine \(\left(f^{-1}\right)'(1)\) for \(f(x)=e^x\).
The answer is \(\displaystyle \left(f^{-1}\right)'(1)=1\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(1)\)
\(\,\,\,\,\,\,1=e^x\)
\(\,\,\,\,\,\,x=0\)
\(\,\,\,\,\,\,f^{-1}(1)=0\)
\(\text{Step 3: Find }f'(0)\)
\(\,\,\,\,\,\,f'(x)=e^x\)
\(\,\,\,\,\,\,f'(0)=1\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(1)=\frac{1}{1}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(1)=1\)
\(\textbf{10)}\) Determine \(\left(f^{-1}\right)'(e^2)\) for \(f(x)=e^x\).
The answer is \(\displaystyle \left(f^{-1}\right)'(e^2)=\frac{1}{e^2}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(e^2)\)
\(\,\,\,\,\,\,e^2=e^x\)
\(\,\,\,\,\,\,x=2\)
\(\,\,\,\,\,\,f^{-1}(e^2)=2\)
\(\text{Step 3: Find }f'(2)\)
\(\,\,\,\,\,\,f'(x)=e^x\)
\(\,\,\,\,\,\,f'(2)=e^2\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(e^2)=\frac{1}{e^2}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(e^2)=\frac{1}{e^2}\)
\(\textbf{11)}\) Determine \(\left(f^{-1}\right)'(3)\) for \(f(x)=\ln{x}\).
The answer is \(\displaystyle \left(f^{-1}\right)'(3)=e^3\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(3)\)
\(\,\,\,\,\,\,3=\ln{x}\)
\(\,\,\,\,\,\,x=e^3\)
\(\,\,\,\,\,\,f^{-1}(3)=e^3\)
\(\text{Step 3: Find }f'(e^3)\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{x}\)
\(\,\,\,\,\,\,f'(e^3)=\frac{1}{e^3}\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(3)=\frac{1}{\frac{1}{e^3}}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(3)=e^3\)
\(\textbf{12)}\) Determine \(\left(f^{-1}\right)'(0)\) for \(f(x)=x^3+x\).
The answer is \(\displaystyle \left(f^{-1}\right)'(0)=1\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(0)\)
\(\,\,\,\,\,\,0=x^3+x\)
\(\,\,\,\,\,\,0=x(x^2+1)\)
\(\,\,\,\,\,\,x=0\)
\(\,\,\,\,\,\,f^{-1}(0)=0\)
\(\text{Step 3: Find }f'(0)\)
\(\,\,\,\,\,\,f'(x)=3x^2+1\)
\(\,\,\,\,\,\,f'(0)=1\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(0)=\frac{1}{1}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(0)=1\)
\(\textbf{13)}\) Determine \(\left(f^{-1}\right)'(5)\) for \(f(x)=x+4\).
The answer is \(\displaystyle \left(f^{-1}\right)'(5)=1\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(5)\)
\(\,\,\,\,\,\,5=x+4\)
\(\,\,\,\,\,\,x=1\)
\(\,\,\,\,\,\,f^{-1}(5)=1\)
\(\text{Step 3: Find }f'(1)\)
\(\,\,\,\,\,\,f'(x)=1\)
\(\,\,\,\,\,\,f'(1)=1\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(5)=\frac{1}{1}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(5)=1\)
\(\textbf{14)}\) Determine \(\left(f^{-1}\right)'(9)\) for \(f(x)=\sqrt{x+7}\).
The answer is \(\displaystyle \left(f^{-1}\right)'(9)=18\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(9)\)
\(\,\,\,\,\,\,9=\sqrt{x+7}\)
\(\,\,\,\,\,\,81=x+7\)
\(\,\,\,\,\,\,x=74\)
\(\,\,\,\,\,\,f^{-1}(9)=74\)
\(\text{Step 3: Find }f'(74)\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x+7}}\)
\(\,\,\,\,\,\,f'(74)=\frac{1}{2\sqrt{81}}=\frac{1}{18}\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(9)=\frac{1}{\frac{1}{18}}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(9)=18\)
\(\textbf{15)}\) Determine \(\left(f^{-1}\right)'(2)\) for \(f(x)=x^3+1\).
The answer is \(\displaystyle \left(f^{-1}\right)'(2)=\frac{1}{3}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(2)\)
\(\,\,\,\,\,\,2=x^3+1\)
\(\,\,\,\,\,\,1=x^3\)
\(\,\,\,\,\,\,x=1\)
\(\,\,\,\,\,\,f^{-1}(2)=1\)
\(\text{Step 3: Find }f'(1)\)
\(\,\,\,\,\,\,f'(x)=3x^2\)
\(\,\,\,\,\,\,f'(1)=3\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(2)=\frac{1}{3}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(2)=\frac{1}{3}\)
\(\textbf{16)}\) Find the derivative of the inverse function of \(f(x)=x^5\) at \(a=32\).
The answer is \(\displaystyle \left(f^{-1}\right)'(32)=\frac{1}{80}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(32)\)
\(\,\,\,\,\,\,32=x^5\)
\(\,\,\,\,\,\,x=2\)
\(\,\,\,\,\,\,f^{-1}(32)=2\)
\(\text{Step 3: Find }f'(2)\)
\(\,\,\,\,\,\,f'(x)=5x^4\)
\(\,\,\,\,\,\,f'(2)=5(2)^4=80\)
\(\text{Step 4: Fill into the formula}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(32)=\frac{1}{80}\)
\(\textbf{17)}\) Find the derivative of the inverse function of \(f(x)=x^3-8\) at \(a=0\).
The answer is \(\displaystyle \left(f^{-1}\right)'(0)=\frac{1}{12}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(0)\)
\(\,\,\,\,\,\,0=x^3-8\)
\(\,\,\,\,\,\,8=x^3\)
\(\,\,\,\,\,\,x=2\)
\(\,\,\,\,\,\,f^{-1}(0)=2\)
\(\text{Step 3: Find }f'(2)\)
\(\,\,\,\,\,\,f'(x)=3x^2\)
\(\,\,\,\,\,\,f'(2)=12\)
\(\text{Step 4: Fill into the formula}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(0)=\frac{1}{12}\)
\(\textbf{18)}\) Find the derivative of the inverse function of \(f(x)=2x+1\) at \(a=9\).
The answer is \(\displaystyle \left(f^{-1}\right)'(9)=\frac{1}{2}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(9)\)
\(\,\,\,\,\,\,9=2x+1\)
\(\,\,\,\,\,\,8=2x\)
\(\,\,\,\,\,\,x=4\)
\(\,\,\,\,\,\,f^{-1}(9)=4\)
\(\text{Step 3: Find }f'(4)\)
\(\,\,\,\,\,\,f'(x)=2\)
\(\,\,\,\,\,\,f'(4)=2\)
\(\text{Step 4: Fill into the formula}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(9)=\frac{1}{2}\)
\(\textbf{19)}\) Find the derivative of the inverse function of \(f(x)=x^2+1,\;x\ge0\) at \(a=10\).
The answer is \(\displaystyle \left(f^{-1}\right)'(10)=\frac{1}{6}\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(10)\)
\(\,\,\,\,\,\,10=x^2+1\)
\(\,\,\,\,\,\,9=x^2\)
\(\,\,\,\,\,\,x=3\left(\text{since }x\ge0\right)\)
\(\,\,\,\,\,\,f^{-1}(10)=3\)
\(\text{Step 3: Find }f'(3)\)
\(\,\,\,\,\,\,f'(x)=2x\)
\(\,\,\,\,\,\,f'(3)=6\)
\(\text{Step 4: Fill into the formula}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(10)=\frac{1}{6}\)
\(\textbf{20)}\) Find the derivative of the inverse function of \(f(x)=\sqrt{x-5}\) at \(a=4\).
The answer is \(\displaystyle \left(f^{-1}\right)'(4)=8\)
\(\text{Step 1: Notes}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(a)=\frac{1}{f’\left(f^{-1}(a)\right)}\)
\(\text{Step 2: Solve for }f^{-1}(4)\)
\(\,\,\,\,\,\,4=\sqrt{x-5}\)
\(\,\,\,\,\,\,16=x-5\)
\(\,\,\,\,\,\,x=21\)
\(\,\,\,\,\,\,f^{-1}(4)=21\)
\(\text{Step 3: Find }f'(21)\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x-5}}\)
\(\,\,\,\,\,\,f'(21)=\frac{1}{2\sqrt{16}}=\frac{1}{8}\)
\(\text{Step 4: Fill into the formula}\)
\(\,\,\,\,\,\,\displaystyle \left(f^{-1}\right)'(4)=\frac{1}{\frac{1}{8}}\)
\(\text{The answer is }\displaystyle \left(f^{-1}\right)'(4)=8\)
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