f
            (
            x
            )
          
        
        =
        
          
            
              log
              
                2
              
            
            ⁡
            (
            x
            )
          
        
        ;
      
    
    {\displaystyle {\color {Blue}f(x)}={\color {Blue}\log _{2}(x)};}
  
 

  
    
      
        
          
            
              f
              ′
            
            (
            x
            )
          
        
        =
        
          
            log
            ⁡
            2
            ⋅
            (
            x
            )
          
        
      
    
    {\displaystyle {\color {Periwinkle}f'(x)}={\color {Periwinkle}\log 2\cdot (x)}}
  


  
    
      
        
          
            
              f
              
                −
                1
              
            
            (
            x
            )
          
        
        =
        
          
            
              2
              
                x
              
            
          
        
        ;
      
    
    {\displaystyle {\color {red}f^{-1}(x)}={\color {red}2^{x}};}
  


  
    
      
        
          
            (
            
              f
              
                −
                1
              
            
            
              )
              ′
            
            (
            x
            )
          
        
        =
        
          
            l
            o
            g
            2
            ⋅
            
              2
              
                x
              
            
          
        
      
    
    {\displaystyle {\color {Salmon}(f^{-1})'(x)}={\color {Salmon}log2\cdot 2^{x}}}
  

விதி:

  
    
      
        
          
            
              f
              ′
            
          
        
        (
        x
        )
        =
        
          
            1
            
              
                
                  (
                  
                    f
                    
                      −
                      1
                    
                  
                  
                    )
                    ′
                  
                
              
              (
              
                
                  f
                
              
              (
              x
              )
              )
            
          
        
      
    
    {\displaystyle {\color {Periwinkle}f'}(x)={\frac {1}{{\color {Salmon}(f^{-1})'}({\color {Blue}f}(x))}}}
  
  
  
    
      
        
          x
          
            0
          
        
        ≈
        5.8
      
    
    {\displaystyle x_{0}\approx 5.8}
  
 எனில்:

  
    
      
        
          
            
              f
              ′
            
          
        
        (
        
          x
          
            0
          
        
        )
        =
        
          
            1
            4
          
        
      
    
    {\displaystyle {\color {Periwinkle}f'}(x_{0})={\frac {1}{4}}}
  


  
    
      
        
          
            (
            
              f
              
                −
                1
              
            
            
              )
              ′
            
          
        
        (
        
          
            f
          
        
        (
        
          x
          
            0
          
        
        )
        )
        =
        4
         
      
    
    {\displaystyle {\color {Salmon}(f^{-1})'}({\color {Blue}f}(x_{0}))=4~}
w:en:Inverse functions and differentiation, w:de:Umkehrregel Shown functions: f ( x ) = log 2 ⁡ x     ;         f − 1 ( x ) = 2 x {\displaystyle {\color {Blue}f(x)=\log _{2}x}~~;~~~~{\color {Red}f^{-1}(x)=2^{x}}} f ′ ( x ) = 1 ln ⁡ 2 ⋅ x     ;         ( f − 1 ) ′ ( x ) = ln ⁡ 2 ⋅ 2 x {\displaystyle {\color {Periwinkle}f'(x)={\frac {1}{\ln 2\cdot x}}}~~;~~~~{\color {Salmon}(f^{-1})'(x)=\ln 2\cdot 2^{x}}} Rule: f ′ ( x ) = 1 ( f − 1 ) ′ ( f ( x ) ) {\displaystyle {\color {Periwinkle}f'}(x)={\frac {1}{{\color {Salmon}(f^{-1})'}({\color {Blue}f}(x))}}} The illustration shows the validity of the rule for the random x 0 = 4 ln ⁡ 2 ≈ 5.77 {\displaystyle x_{0}={\frac {4}{\ln 2}}\approx 5.77} : f ′ ( x 0 ) = 1 4 {\displaystyle {\color {Periwinkle}f'}(x_{0})={\frac {1}{4}}} ( f − 1 ) ′ ( f ( x 0 ) ) = 4   {\displaystyle {\color {Salmon}(f^{-1})'}({\color {Blue}f}(x_{0}))=4~}